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Title: Amusements in Mathematics

Author: Henry Ernest Dudeney

Release date: September 17, 2005 [eBook #16713]
Most recently updated: December 12, 2020

Language: English

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*** START OF THE PROJECT GUTENBERG EBOOK AMUSEMENTS
IN MATHEMATICS ***

Transcribers note: Many of the puzzles in this book assume a
familiarity with the currency of Great Britain in the early 1900s.
As this is likely not common knowledge for those outside Britain

(and possibly many within,) I am including a chart of relative
values.
The most common units used were:

the Penny, abbreviated: d. (from the Roman penny, denarius)
the Shilling, abbreviated: s.
the Pound, abbreviated: £

There was 12 Pennies to a Shilling and 20 Shillings to a Pound,
so there was 240 Pennies in a Pound.
To further complicate things, there were many coins which were
various fractional values of Pennies, Shillings or Pounds.

Farthing ¼d.
Half-penny ½d.
Penny 1d.
Three-penny 3d.
Sixpence (or tanner) 6d.
Shilling (or bob) 1s.
Florin or two shilling piece 2s.
Half-crown (or half-dollar) 2s. 6d.
Double-florin 4s.
Crown (or dollar) 5s.
Half-Sovereign 10s.
Sovereign (or Pound) £1 or 20s.

This is by no means a comprehensive list, but it should be
adequate to solve the puzzles in this book.

AMUSEMENTS IN
MATHEMATICS
by

HENRY ERNEST DUDENEY

In Mathematicks he was greater
Than Tycho Brahe or Erra Pater:
For he, by geometrick scale,
Could take the size of pots of ale;
Resolve, by sines and tangents,
straight,
If bread or butter wanted weight;
And wisely tell what hour o' th'
day
The clock does strike by algebra.

BUTLER'S
Hudib
ras.

1917

PREFACE
In issuing this volume of my Mathematical Puzzles, of which some have
appeared in periodicals and others are given here for the first time, I must
acknowledge the encouragement that I have received from many unknown
correspondents, at home and abroad, who have expressed a desire to have
the problems in a collected form, with some of the solutions given at greater
length than is possible in magazines and newspapers. Though I have
included a few old puzzles that have interested the world for generations,
where I felt that there was something new to be said about them, the
problems are in the main original. It is true that some of these have become
widely known through the press, and it is possible that the reader may be
glad to know their source.
On the question of Mathematical Puzzles in general there is, perhaps, little
more to be said than I have written elsewhere. The history of the subject
entails nothing short of the actual story of the beginnings and development
of exact thinking in man. The historian must start from the time when man
first succeeded in counting his ten fingers and in dividing an apple into two
approximately equal parts. Every puzzle that is worthy of consideration can
be referred to mathematics and logic. Every man, woman, and child who
tries to "reason out" the answer to the simplest puzzle is working, though
not of necessity consciously, on mathematical lines. Even those puzzles that
we have no way of attacking except by haphazard attempts can be brought
under a method of what has been called "glorified trial"—a system of
shortening our labours by avoiding or eliminating what our reason tells us is
useless. It is, in fact, not easy to say sometimes where the "empirical"
begins and where it ends.
When a man says, "I have never solved a puzzle in my life," it is difficult to
know exactly what he means, for every intelligent individual is doing it
every day. The unfortunate inmates of our lunatic asylums are sent there
expressly because they cannot solve puzzles—because they have lost their
powers of reason. If there were no puzzles to solve, there would be no
questions to ask; and if there were no questions to be asked, what a world it

would be! We should all be equally omniscient, and conversation would be
useless and idle.
It is possible that some few exceedingly sober-minded mathematicians, who
are impatient of any terminology in their favourite science but the
academic, and who object to the elusive x and y appearing under any other
names, will have wished that various problems had been presented in a less
popular dress and introduced with a less flippant phraseology. I can only
refer them to the first word of my title and remind them that we are
primarily out to be amused—not, it is true, without some hope of picking up
morsels of knowledge by the way. If the manner is light, I can only say, in
the words of Touchstone, that it is "an ill-favoured thing, sir, but my own; a
poor humour of mine, sir."
As for the question of difficulty, some of the puzzles, especially in the
Arithmetical and Algebraical category, are quite easy. Yet some of those
examples that look the simplest should not be passed over without a little
consideration, for now and again it will be found that there is some more or
less subtle pitfall or trap into which the reader may be apt to fall. It is good
exercise to cultivate the habit of being very wary over the exact wording of
a puzzle. It teaches exactitude and caution. But some of the problems are
very hard nuts indeed, and not unworthy of the attention of the advanced
mathematician. Readers will doubtless select according to their individual
tastes.
In many cases only the mere answers are given. This leaves the beginner
something to do on his own behalf in working out the method of solution,
and saves space that would be wasted from the point of view of the
advanced student. On the other hand, in particular cases where it seemed
likely to interest, I have given rather extensive solutions and treated
problems in a general manner. It will often be found that the notes on one
problem will serve to elucidate a good many others in the book; so that the
reader's difficulties will sometimes be found cleared up as he advances.
Where it is possible to say a thing in a manner that may be "understanded of
the people" generally, I prefer to use this simple phraseology, and so engage
the attention and interest of a larger public. The mathematician will in such
cases have no difficulty in expressing the matter under consideration in
terms of his familiar symbols.

I have taken the greatest care in reading the proofs, and trust that any errors
that may have crept in are very few. If any such should occur, I can only
plead, in the words of Horace, that "good Homer sometimes nods," or, as
the bishop put it, "Not even the youngest curate in my diocese is infallible."
I have to express my thanks in particular to the proprietors of The Strand
Magazine, Cassell's Magazine, The Queen, Tit-Bits, and The Weekly
Dispatch for their courtesy in allowing me to reprint some of the puzzles
that have appeared in their pages.
THE AUTHORS' CLUB
March 25, 1917

CONTENTS
PREFACE v
ARITHMETICAL AND ALGEBRAICAL PROBLEMS. 1
Money Puzzles. 1
Age and Kinship Puzzles. 6
Clock Puzzles. 9
Locomotion and Speed Puzzles. 11
Digital Puzzles. 13
Various Arithmetical and Algebraical Problems. 17
GEOMETRICAL PROBLEMS. 27
Dissection Puzzles. 27
Greek Cross Puzzles. 28
Various Dissection Puzzles. 35
Patchwork Puzzles 46
Various Geometrical Puzzles. 49
POINTS AND LINES PROBLEMS. 56
MOVING COUNTER PROBLEMS. 58
UNICURSAL AND ROUTE PROBLEMS. 68
COMBINATION AND GROUP PROBLEMS. 76
CHESSBOARD PROBLEMS. 85
The Chessboard. 85
Statical Chess Puzzles. 88
The Guarded Chessboard. 95
Dynamical Chess Puzzles. 96
Various Chess Puzzles. 112
MEASURING, WEIGHING, AND PACKING PUZZLES. 109
CROSSING RIVER PROBLEMS 112
PROBLEMS CONCERNING GAMES. 114
PUZZLE GAMES. 117
MAGIC SQUARE PROBLEMS. 119
Subtracting, Multiplying, and Dividing Magics. 124

Magic Squares of Primes. 125
MAZES AND HOW TO THREAD THEM. 127
THE PARADOX PARTY. 137
UNCLASSIFIED PROBLEMS. 142
SOLUTIONS. 148
INDEX. 253

AMUSEMENTS IN
MATHEMATICS.

ARITHMETICAL AND ALGEBRAICAL
PROBLEMS.
"And what was he?
Forsooth, a great arithmetician."
Othello, I. i.
The puzzles in this department are roughly thrown together in classes for
the convenience of the reader. Some are very easy, others quite difficult.
But they are not arranged in any order of difficulty—and this is intentional,
for it is well that the solver should not be warned that a puzzle is just what it
seems to be. It may, therefore, prove to be quite as simple as it looks, or it
may contain some pitfall into which, through want of care or over-
confidence, we may stumble.
Also, the arithmetical and algebraical puzzles are not separated in the
manner adopted by some authors, who arbitrarily require certain problems
to be solved by one method or the other. The reader is left to make his own
choice and determine which puzzles are capable of being solved by him on
purely arithmetical lines.

MONEY PUZZLES.
"Put not your trust in money, but put your money in trust."
OLIVER WENDELL HOLMES.

1.—A POST-OFFICE PERPLEXITY.
In every business of life we are occasionally perplexed by some chance
question that for the moment staggers us. I quite pitied a young lady in a
branch post-office when a gentleman entered and deposited a crown on the
counter with this request: "Please give me some twopenny stamps, six times
as many penny stamps, and make up the rest of the money in twopence-
halfpenny stamps." For a moment she seemed bewildered, then her brain
cleared, and with a smile she handed over stamps in exact fulfilment of the
order. How long would it have taken you to think it out?

2.—YOUTHFUL PRECOCITY.
The precocity of some youths is surprising. One is disposed to say on
occasion, "That boy of yours is a genius, and he is certain to do great things
when he grows up;" but past experience has taught us that he invariably
becomes quite an ordinary citizen. It is so often the case, on the contrary,
that the dull boy becomes a great man. You never can tell. Nature loves to
present to us these queer paradoxes. It is well known that those wonderful
"lightning calculators," who now and again surprise the world by their feats,
lose all their mysterious powers directly they are taught the elementary
rules of arithmetic.
A boy who was demolishing a choice banana was approached by a young
friend, who, regarding him with envious eyes, asked, "How much did you

pay for that banana, Fred?" The prompt answer was quite remarkable in its
way: "The man what I bought it of receives just half as many sixpences for
sixteen dozen dozen bananas as he gives bananas for a fiver."
Now, how long will it take the reader to say correctly just how much Fred
paid for his rare and refreshing fruit?

3.—AT A CATTLE MARKET.
Three countrymen met at a cattle market. "Look here," said Hodge to Jakes,
"I'll give you six of my pigs for one of your horses, and then you'll have
twice as many animals here as I've got." "If that's your way of doing
business," said Durrant to Hodge, "I'll give you fourteen of my sheep for a
horse, and then you'll have three times as many animals as I." "Well, I'll go
better than that," said Jakes to Durrant; "I'll give you four cows for a horse,
and then you'll have six times as many animals as I've got here."
No doubt this was a very primitive way of bartering animals, but it is an
interesting little puzzle to discover just how many animals Jakes, Hodge,
and Durrant must have taken to the cattle market.

4.—THE BEANFEAST PUZZLE.
A number of men went out together on a bean-feast. There were four parties
invited—namely, 25 cobblers, 20 tailors, 18 hatters, and 12 glovers. They
spent altogether £6, 13s. It was found that five cobblers spent as much as
four tailors; that twelve tailors spent as much as nine hatters; and that six
hatters spent as much as eight glovers. The puzzle is to find out how much
each of the four parties spent.

5.—A QUEER COINCIDENCE.

Seven men, whose names were Adams, Baker, Carter, Dobson, Edwards,
Francis, and Gudgeon, were recently engaged in play. The name of the
particular game is of no consequence. They had agreed that whenever a
player won a game he should double the money of each of the other players
—that is, he was to give the players just as much money as they had already
in their pockets. They played seven games, and, strange to say, each won a
game in turn, in the order in which their names are given. But a more
curious coincidence is this—that when they had finished play each of the
seven men had exactly the same amount—two shillings and eightpence—in
his pocket. The puzzle is to find out how much money each man had with
him before he sat down to play.

6.—A CHARITABLE BEQUEST.
A man left instructions to his executors to distribute once a year exactly
fifty-five shillings among the poor of his parish; but they were only to
continue the gift so long as they could make it in different ways, always
giving eighteenpence each to a number of women and half a crown each to
men. During how many years could the charity be administered? Of course,
by "different ways" is meant a different number of men and women every
time.

7.—THE WIDOW'S LEGACY.
A gentleman who recently died left the sum of £8,000 to be divided among
his widow, five sons, and four daughters. He directed that every son should
receive three times as much as a daughter, and that every daughter should
have twice as much as their mother. What was the widow's share?

8.—INDISCRIMINATE CHARITY.

A charitable gentleman, on his way home one night, was appealed to by
three needy persons in succession for assistance. To the first person he gave
one penny more than half the money he had in his pocket; to the second
person he gave twopence more than half the money he then had in his
pocket; and to the third person he handed over threepence more than half of
what he had left. On entering his house he had only one penny in his
pocket. Now, can you say exactly how much money that gentleman had on
him when he started for home?

9.—THE TWO AEROPLANES.
A man recently bought two aeroplanes, but afterwards found that they
would not answer the purpose for which he wanted them. So he sold them
for £600 each, making a loss of 20 per cent, on one machine and a profit of
20 per cent, on the other. Did he make a profit on the whole transaction, or a
loss? And how much?

10.—BUYING PRESENTS.
"Whom do you think I met in town last week, Brother William?" said Uncle
Benjamin. "That old skinflint Jorkins. His family had been taking him
around buying Christmas presents. He said to me, 'Why cannot the
government abolish Christmas, and make the giving of presents punishable
by law? I came out this morning with a certain amount of money in my
pocket, and I find I have spent just half of it. In fact, if you will believe me,
I take home just as many shillings as I had pounds, and half as many
pounds as I had shillings. It is monstrous!'" Can you say exactly how much
money Jorkins had spent on those presents?

11.—THE CYCLISTS' FEAST.

'Twas last Bank Holiday, so I've been told,
Some cyclists rode abroad in glorious weather.
Resting at noon within a tavern old,
They all agreed to have a feast together.
"Put it all in one bill, mine host," they said,
"For every man an equal share will pay."
The bill was promptly on the table laid,
And four pounds was the reckoning that day.
But, sad to state, when they prepared to square,
'Twas found that two had sneaked outside and fled.
So, for two shillings more than his due share
Each honest man who had remained was bled.
They settled later with those rogues, no doubt.
How many were they when they first set out?

12.—A QUEER THING IN MONEY.
It will be found that £66, 6s. 6d. equals 15,918 pence. Now, the four 6's
added together make 24, and the figures in 15,918 also add to 24. It is a
curious fact that there is only one other sum of money, in pounds, shillings,
and pence (all similarly repetitions of one figure), of which the digits shall
add up the same as the digits of the amount in pence. What is the other sum
of money?

13.—A NEW MONEY PUZZLE.
The largest sum of money that can be written in pounds, shillings, pence,
and farthings, using each of the nine digits once and only once, is £98,765,
4s. 3½d. Now, try to discover the smallest sum of money that can be written
down under precisely the same conditions. There must be some value given
for each denomination—pounds, shillings, pence, and farthings—and the
nought may not be used. It requires just a little judgment and thought.

14.—SQUARE MONEY.
"This is queer," said McCrank to his friend. "Twopence added to twopence
is fourpence, and twopence multiplied by twopence is also fourpence." Of
course, he was wrong in thinking you can multiply money by money. The
multiplier must be regarded as an abstract number. It is true that two feet
multiplied by two feet will make four square feet. Similarly, two pence
multiplied by two pence will produce four square pence! And it will perplex
the reader to say what a "square penny" is. But we will assume for the
purposes of our puzzle that twopence multiplied by twopence is fourpence.
Now, what two amounts of money will produce the next smallest possible
result, the same in both cases, when added or multiplied in this manner?
The two amounts need not be alike, but they must be those that can be paid
in current coins of the realm.

15.—POCKET MONEY.
What is the largest sum of money—all in current silver coins and no four-
shilling piece—that I could have in my pocket without being able to give
change for a half-sovereign?

16.—THE MILLIONAIRE'S PERPLEXITY.
Mr. Morgan G. Bloomgarten, the millionaire, known in the States as the
Clam King, had, for his sins, more money than he knew what to do with. It
bored him. So he determined to persecute some of his poor but happy
friends with it. They had never done him any harm, but he resolved to
inoculate them with the "source of all evil." He therefore proposed to
distribute a million dollars among them and watch them go rapidly to the
bad. But he was a man of strange fancies and superstitions, and it was an
inviolable rule with him never to make a gift that was not either one dollar

or some power of seven—such as 7, 49, 343, 2,401, which numbers of
dollars are produced by simply multiplying sevens together. Another rule of
his was that he would never give more than six persons exactly the same
sum. Now, how was he to distribute the 1,000,000 dollars? You may
distribute the money among as many people as you like, under the
conditions given.

17.—THE PUZZLING MONEY-BOXES.
Four brothers—named John, William, Charles, and Thomas—had each a
money-box. The boxes were all given to them on the same day, and they at
once put what money they had into them; only, as the boxes were not very
large, they first changed the money into as few coins as possible. After they
had done this, they told one another how much money they had saved, and
it was found that if John had had 2s. more in his box than at present, if
William had had 2s. less, if Charles had had twice as much, and if Thomas
had had half as much, they would all have had exactly the same amount.
Now, when I add that all four boxes together contained 45s., and that there
were only six coins in all in them, it becomes an entertaining puzzle to
discover just what coins were in each box.

18.—THE MARKET WOMEN.
A number of market women sold their various products at a certain price
per pound (different in every case), and each received the same amount—
2s. 2½d. What is the greatest number of women there could have been? The
price per pound in every case must be such as could be paid in current
money.

19.—THE NEW YEAR'S EVE SUPPERS.
The proprietor of a small London café has given me some interesting
figures. He says that the ladies who come alone to his place for refreshment
spend each on an average eighteenpence, that the unaccompanied men
spend half a crown each, and that when a gentleman brings in a lady he
spends half a guinea. On New Year's Eve he supplied suppers to twenty-five
persons, and took five pounds in all. Now, assuming his averages to have
held good in every case, how was his company made up on that occasion?
Of course, only single gentlemen, single ladies, and pairs (a lady and
gentleman) can be supposed to have been present, as we are not considering
larger parties.

20.—BEEF AND SAUSAGES.
"A neighbour of mine," said Aunt Jane, "bought a certain quantity of beef at
two shillings a pound, and the same quantity of sausages at eighteenpence a
pound. I pointed out to her that if she had divided the same money equally
between beef and sausages she would have gained two pounds in the total
weight. Can you tell me exactly how much she spent?"
"Of course, it is no business of mine," said Mrs. Sunniborne; "but a lady
who could pay such prices must be somewhat inexperienced in domestic
economy."
"I quite agree, my dear," Aunt Jane replied, "but you see that is not the
precise point under discussion, any more than the name and morals of the
tradesman."

21.—A DEAL IN APPLES.
I paid a man a shilling for some apples, but they were so small that I made
him throw in two extra apples. I find that made them cost just a penny a

dozen less than the first price he asked. How many apples did I get for my
shilling?

22.—A DEAL IN EGGS.
A man went recently into a dairyman's shop to buy eggs. He wanted them of
various qualities. The salesman had new-laid eggs at the high price of
fivepence each, fresh eggs at one penny each, eggs at a halfpenny each, and
eggs for electioneering purposes at a greatly reduced figure, but as there
was no election on at the time the buyer had no use for the last. However,
he bought some of each of the three other kinds and obtained exactly one
hundred eggs for eight and fourpence. Now, as he brought away exactly the
same number of eggs of two of the three qualities, it is an interesting puzzle
to determine just how many he bought at each price.

23.—THE CHRISTMAS-BOXES.
Some years ago a man told me he had spent one hundred English silver
coins in Christmas-boxes, giving every person the same amount, and it cost
him exactly £1, 10s. 1d. Can you tell just how many persons received the
present, and how he could have managed the distribution? That odd penny
looks queer, but it is all right.

24.—A SHOPPING PERPLEXITY.
Two ladies went into a shop where, through some curious eccentricity, no
change was given, and made purchases amounting together to less than five
shillings. "Do you know," said one lady, "I find I shall require no fewer than
six current coins of the realm to pay for what I have bought." The other lady
considered a moment, and then exclaimed: "By a peculiar coincidence, I am

exactly in the same dilemma." "Then we will pay the two bills together."
But, to their astonishment, they still required six coins. What is the smallest
possible amount of their purchases—both different?

25.—CHINESE MONEY.
The Chinese are a curious people, and have strange inverted ways of doing
things. It is said that they use a saw with an upward pressure instead of a
downward one, that they plane a deal board by pulling the tool toward them
instead of pushing it, and that in building a house they first construct the
roof and, having raised that into position, proceed to work downwards. In
money the currency of the country consists of taels of fluctuating value. The
tael became thinner and thinner until 2,000 of them piled together made less
than three inches in height. The common cash consists of brass coins of
varying thicknesses, with a round, square, or triangular hole in the centre, as
in our illustration.

These are strung on wires like buttons. Supposing that eleven coins with
round holes are worth fifteen ching-changs, that eleven with square holes
are worth sixteen ching-changs, and that eleven with triangular holes are
worth seventeen ching-changs, how can a Chinaman give me change for
half a crown, using no coins other than the three mentioned? A ching-chang
is worth exactly twopence and four-fifteenths of a ching-chang.

26.—THE JUNIOR CLERK'S PUZZLE.

Two youths, bearing the pleasant names of Moggs and Snoggs, were
employed as junior clerks by a merchant in Mincing Lane. They were both
engaged at the same salary—that is, commencing at the rate of £50 a year,
payable half-yearly. Moggs had a yearly rise of £10, and Snoggs was
offered the same, only he asked, for reasons that do not concern our puzzle,
that he might take his rise at £2, 10s. half-yearly, to which his employer
(not, perhaps, unnaturally!) had no objection.
Now we come to the real point of the puzzle. Moggs put regularly into the
Post Office Savings Bank a certain proportion of his salary, while Snoggs
saved twice as great a proportion of his, and at the end of five years they
had together saved £268, 15s. How much had each saved? The question of
interest can be ignored.

27.—GIVING CHANGE.
Every one is familiar with the difficulties that frequently arise over the
giving of change, and how the assistance of a third person with a few coins
in his pocket will sometimes help us to set the matter right. Here is an
example. An Englishman went into a shop in New York and bought goods
at a cost of thirty-four cents. The only money he had was a dollar, a three-
cent piece, and a two-cent piece. The tradesman had only a half-dollar and a
quarter-dollar. But another customer happened to be present, and when
asked to help produced two dimes, a five-cent piece, a two-cent piece, and a
one-cent piece. How did the tradesman manage to give change? For the
benefit of those readers who are not familiar with the American coinage, it
is only necessary to say that a dollar is a hundred cents and a dime ten
cents. A puzzle of this kind should rarely cause any difficulty if attacked in
a proper manner.

28.—DEFECTIVE OBSERVATION.

Our observation of little things is frequently defective, and our memories
very liable to lapse. A certain judge recently remarked in a case that he had
no recollection whatever of putting the wedding-ring on his wife's finger.
Can you correctly answer these questions without having the coins in sight?
On which side of a penny is the date given? Some people are so
unobservant that, although they are handling the coin nearly every day of
their lives, they are at a loss to answer this simple question. If I lay a penny
flat on the table, how many other pennies can I place around it, every one
also lying flat on the table, so that they all touch the first one? The
geometrician will, of course, give the answer at once, and not need to make
any experiment. He will also know that, since all circles are similar, the
same answer will necessarily apply to any coin. The next question is a most
interesting one to ask a company, each person writing down his answer on a
slip of paper, so that no one shall be helped by the answers of others. What
is the greatest number of three-penny-pieces that may be laid flat on the
surface of a half-crown, so that no piece lies on another or overlaps the
surface of the half-crown? It is amazing what a variety of different answers
one gets to this question. Very few people will be found to give the correct
number. Of course the answer must be given without looking at the coins.

29.—THE BROKEN COINS.
A man had three coins—a sovereign, a shilling, and a penny—and he found
that exactly the same fraction of each coin had been broken away. Now,
assuming that the original intrinsic value of these coins was the same as
their nominal value—that is, that the sovereign was worth a pound, the
shilling worth a shilling, and the penny worth a penny—what proportion of
each coin has been lost if the value of the three remaining fragments is
exactly one pound?

30.—TWO QUESTIONS IN PROBABILITIES.

There is perhaps no class of puzzle over which people so frequently blunder
as that which involves what is called the theory of probabilities. I will give
two simple examples of the sort of puzzle I mean. They are really quite
easy, and yet many persons are tripped up by them. A friend recently
produced five pennies and said to me: "In throwing these five pennies at the
same time, what are the chances that at least four of the coins will turn up
either all heads or all tails?" His own solution was quite wrong, but the
correct answer ought not to be hard to discover. Another person got a wrong
answer to the following little puzzle which I heard him propound: "A man
placed three sovereigns and one shilling in a bag. How much should be paid
for permission to draw one coin from it?" It is, of course, understood that
you are as likely to draw any one of the four coins as another.

31.—DOMESTIC ECONOMY.
Young Mrs. Perkins, of Putney, writes to me as follows: "I should be very
glad if you could give me the answer to a little sum that has been worrying
me a good deal lately. Here it is: We have only been married a short time,
and now, at the end of two years from the time when we set up
housekeeping, my husband tells me that he finds we have spent a third of
his yearly income in rent, rates, and taxes, one-half in domestic expenses,
and one-ninth in other ways. He has a balance of £190 remaining in the
bank. I know this last, because he accidentally left out his pass-book the
other day, and I peeped into it. Don't you think that a husband ought to give
his wife his entire confidence in his money matters? Well, I do; and—will
you believe it?—he has never told me what his income really is, and I want,
very naturally, to find out. Can you tell me what it is from the figures I have
given you?"
Yes; the answer can certainly be given from the figures contained in Mrs.
Perkins's letter. And my readers, if not warned, will be practically
unanimous in declaring the income to be—something absurdly in excess of
the correct answer!

32.—THE EXCURSION TICKET PUZZLE.
When the big flaming placards were exhibited at the little provincial
railway station, announcing that the Great —— Company would run cheap
excursion trains to London for the Christmas holidays, the inhabitants of
Mudley-cum-Turmits were in quite a flutter of excitement. Half an hour
before the train came in the little booking office was crowded with country
passengers, all bent on visiting their friends in the great Metropolis. The
booking clerk was unaccustomed to dealing with crowds of such a
dimension, and he told me afterwards, while wiping his manly brow, that
what caused him so much trouble was the fact that these rustics paid their
fares in such a lot of small money.
He said that he had enough farthings to supply a West End draper with
change for a week, and a sufficient number of threepenny pieces for the
congregations of three parish churches. "That excursion fare," said he, "is
nineteen shillings and ninepence, and I should like to know in just how
many different ways it is possible for such an amount to be paid in the
current coin of this realm."
Here, then, is a puzzle: In how many different ways may nineteen shillings
and ninepence be paid in our current coin? Remember that the fourpenny-
piece is not now current.

33.—A PUZZLE IN REVERSALS.
Most people know that if you take any sum of money in pounds, shillings,
and pence, in which the number of pounds (less than £12) exceeds that of
the pence, reverse it (calling the pounds pence and the pence pounds), find
the difference, then reverse and add this difference, the result is always £12,
18s. 11d. But if we omit the condition, "less than £12," and allow nought to
represent shillings or pence—(1) What is the lowest amount to which the
rule will not apply? (2) What is the highest amount to which it will apply?
Of course, when reversing such a sum as £14, 15s. 3d. it may be written £3,
16s. 2d., which is the same as £3, 15s. 14d.

34.—THE GROCER AND DRAPER.
A country "grocer and draper" had two rival assistants, who prided
themselves on their rapidity in serving customers. The young man on the
grocery side could weigh up two one-pound parcels of sugar per minute,
while the drapery assistant could cut three one-yard lengths of cloth in the
same time. Their employer, one slack day, set them a race, giving the grocer
a barrel of sugar and telling him to weigh up forty-eight one-pound parcels
of sugar While the draper divided a roll of forty-eight yards of cloth into
yard pieces. The two men were interrupted together by customers for nine
minutes, but the draper was disturbed seventeen times as long as the grocer.
What was the result of the race?

35.—JUDKINS'S CATTLE.
Hiram B. Judkins, a cattle-dealer of Texas, had five droves of animals,
consisting of oxen, pigs, and sheep, with the same number of animals in
each drove. One morning he sold all that he had to eight dealers. Each
dealer bought the same number of animals, paying seventeen dollars for
each ox, four dollars for each pig, and two dollars for each sheep; and
Hiram received in all three hundred and one dollars. What is the greatest
number of animals he could have had? And how many would there be of
each kind?

36.—BUYING APPLES.
As the purchase of apples in small quantities has always presented
considerable difficulties, I think it well to offer a few remarks on this
subject. We all know the story of the smart boy who, on being told by the
old woman that she was selling her apples at four for threepence, said: "Let

me see! Four for threepence; that's three for twopence, two for a penny, one
for nothing—I'll take one!"
There are similar cases of perplexity. For example, a boy once picked up a
penny apple from a stall, but when he learnt that the woman's pears were
the same price he exchanged it, and was about to walk off. "Stop!" said the
woman. "You haven't paid me for the pear!" "No," said the boy, "of course
not. I gave you the apple for it." "But you didn't pay for the apple!" "Bless
the woman! You don't expect me to pay for the apple and the pear too!" And
before the poor creature could get out of the tangle the boy had disappeared.
Then, again, we have the case of the man who gave a boy sixpence and
promised to repeat the gift as soon as the youngster had made it into
ninepence. Five minutes later the boy returned. "I have made it into
ninepence," he said, at the same time handing his benefactor threepence.
"How do you make that out?" he was asked. "I bought threepennyworth of
apples." "But that does not make it into ninepence!" "I should rather think it
did," was the boy's reply. "The apple woman has threepence, hasn't she?
Very well, I have threepennyworth of apples, and I have just given you the
other threepence. What's that but ninepence?"
I cite these cases just to show that the small boy really stands in need of a
little instruction in the art of buying apples. So I will give a simple poser
dealing with this branch of commerce.
An old woman had apples of three sizes for sale—one a penny, two a penny,
and three a penny. Of course two of the second size and three of the third
size were respectively equal to one apple of the largest size. Now, a
gentleman who had an equal number of boys and girls gave his children
sevenpence to be spent amongst them all on these apples. The puzzle is to
give each child an equal distribution of apples. How was the sevenpence
spent, and how many children were there?

37.—BUYING CHESTNUTS.

Though the following little puzzle deals with the purchase of chestnuts, it is
not itself of the "chestnut" type. It is quite new. At first sight it has certainly
the appearance of being of the "nonsense puzzle" character, but it is all right
when properly considered.
A man went to a shop to buy chestnuts. He said he wanted a pennyworth,
and was given five chestnuts. "It is not enough; I ought to have a sixth," he
remarked! "But if I give you one chestnut more." the shopman replied, "you
will have five too many." Now, strange to say, they were both right. How
many chestnuts should the buyer receive for half a crown?

38.—THE BICYCLE THIEF.
Here is a little tangle that is perpetually cropping up in various guises. A
cyclist bought a bicycle for £15 and gave in payment a cheque for £25. The
seller went to a neighbouring shopkeeper and got him to change the cheque
for him, and the cyclist, having received his £10 change, mounted the
machine and disappeared. The cheque proved to be valueless, and the
salesman was requested by his neighbour to refund the amount he had
received. To do this, he was compelled to borrow the £25 from a friend, as
the cyclist forgot to leave his address, and could not be found. Now, as the
bicycle cost the salesman £11, how much money did he lose altogether?

39.—THE COSTERMONGER'S PUZZLE.
"How much did yer pay for them oranges, Bill?"
"I ain't a-goin' to tell yer, Jim. But I beat the old cove down fourpence a
hundred."
"What good did that do yer?"
"Well, it meant five more oranges on every ten shillin's-worth."

Now, what price did Bill actually pay for the oranges? There is only one
rate that will fit in with his statements.

AGE AND KINSHIP PUZZLES.
"The days of our years are threescore years and ten."
—Psalm xc. 10.
For centuries it has been a favourite method of propounding arithmetical
puzzles to pose them in the form of questions as to the age of an individual.
They generally lend themselves to very easy solution by the use of algebra,
though often the difficulty lies in stating them correctly. They may be made
very complex and may demand considerable ingenuity, but no general laws
can well be laid down for their solution. The solver must use his own
sagacity. As for puzzles in relationship or kinship, it is quite curious how
bewildering many people find these things. Even in ordinary conversation,
some statement as to relationship, which is quite clear in the mind of the
speaker, will immediately tie the brains of other people into knots. Such
expressions as "He is my uncle's son-in-law's sister" convey absolutely
nothing to some people without a detailed and laboured explanation. In
such cases the best course is to sketch a brief genealogical table, when the
eye comes immediately to the assistance of the brain. In these days, when
we have a growing lack of respect for pedigrees, most people have got out
of the habit of rapidly drawing such tables, which is to be regretted, as they
would save a lot of time and brain racking on occasions.

40.—MAMMA'S AGE.
Tommy: "How old are you, mamma?"
Mamma: "Let me think, Tommy. Well, our three ages add up to exactly
seventy years."
Tommy: "That's a lot, isn't it? And how old are you, papa?"
Papa: "Just six times as old as you, my son."

Tommy: "Shall I ever be half as old as you, papa?"
Papa: "Yes, Tommy; and when that happens our three ages will add up to
exactly twice as much as to-day."
Tommy: "And supposing I was born before you, papa; and supposing
mamma had forgot all about it, and hadn't been at home when I came; and
supposing——"
Mamma: "Supposing, Tommy, we talk about bed. Come along, darling.
You'll have a headache."
Now, if Tommy had been some years older he might have calculated the
exact ages of his parents from the information they had given him. Can you
find out the exact age of mamma?

41.—THEIR AGES.
"My husband's age," remarked a lady the other day, "is represented by the
figures of my own age reversed. He is my senior, and the difference
between our ages is one-eleventh of their sum."

42.—THE FAMILY AGES.
When the Smileys recently received a visit from the favourite uncle, the
fond parents had all the five children brought into his presence. First came
Billie and little Gertrude, and the uncle was informed that the boy was
exactly twice as old as the girl. Then Henrietta arrived, and it was pointed
out that the combined ages of herself and Gertrude equalled twice the age of
Billie. Then Charlie came running in, and somebody remarked that now the
combined ages of the two boys were exactly twice the combined ages of the
two girls. The uncle was expressing his astonishment at these coincidences
when Janet came in. "Ah! uncle," she exclaimed, "you have actually arrived
on my twenty-first birthday!" To this Mr. Smiley added the final staggerer:

"Yes, and now the combined ages of the three girls are exactly equal to
twice the combined ages of the two boys." Can you give the age of each
child?

43.—MRS. TIMPKINS'S AGE.
Edwin: "Do you know, when the Timpkinses married eighteen years ago
Timpkins was three times as old as his wife, and to-day he is just twice as
old as she?"
Angelina: "Then how old was Mrs. Timpkins on the wedding day?"
Can you answer Angelina's question?

44—A CENSUS PUZZLE.
Mr. and Mrs. Jorkins have fifteen children, all born at intervals of one year
and a half. Miss Ada Jorkins, the eldest, had an objection to state her age to
the census man, but she admitted that she was just seven times older than
little Johnnie, the youngest of all. What was Ada's age? Do not too hastily
assume that you have solved this little poser. You may find that you have
made a bad blunder!

45.—MOTHER AND DAUGHTER.
"Mother, I wish you would give me a bicycle," said a girl of twelve the
other day.
"I do not think you are old enough yet, my dear," was the reply. "When I am
only three times as old as you are you shall have one."

Now, the mother's age is forty-five years. When may the young lady expect
to receive her present?

46.—MARY AND MARMADUKE.
Marmaduke: "Do you know, dear, that in seven years' time our combined
ages will be sixty-three years?"
Mary: "Is that really so? And yet it is a fact that when you were my present
age you were twice as old as I was then. I worked it out last night."
Now, what are the ages of Mary and Marmaduke?

47—ROVER'S AGE.
"Now, then, Tommy, how old is Rover?" Mildred's young man asked her
brother.
"Well, five years ago," was the youngster's reply, "sister was four times
older than the dog, but now she is only three times as old."
Can you tell Rover's age?

48.—CONCERNING TOMMY'S AGE.
Tommy Smart was recently sent to a new school. On the first day of his
arrival the teacher asked him his age, and this was his curious reply: "Well,
you see, it is like this. At the time I was born—I forget the year—my only
sister, Ann, happened to be just one-quarter the age of mother, and she is
now one-third the age of father." "That's all very well," said the teacher,
"but what I want is not the age of your sister Ann, but your own age." "I
was just coming to that," Tommy answered; "I am just a quarter of mother's

present age, and in four years' time I shall be a quarter the age of father.
Isn't that funny?"
This was all the information that the teacher could get out of Tommy Smart.
Could you have told, from these facts, what was his precise age? It is
certainly a little puzzling.

49.—NEXT-DOOR NEIGHBOURS.
There were two families living next door to one another at Tooting Bec—
the Jupps and the Simkins. The united ages of the four Jupps amounted to
one hundred years, and the united ages of the Simkins also amounted to the
same. It was found in the case of each family that the sum obtained by
adding the squares of each of the children's ages to the square of the
mother's age equalled the square of the father's age. In the case of the Jupps,
however, Julia was one year older than her brother Joe, whereas Sophy
Simkin was two years older than her brother Sammy. What was the age of
each of the eight individuals?

50.—THE BAG OF NUTS.
Three boys were given a bag of nuts as a Christmas present, and it was
agreed that they should be divided in proportion to their ages, which
together amounted to 17½ years. Now the bag contained 770 nuts, and as
often as Herbert took four Robert took three, and as often as Herbert took
six Christopher took seven. The puzzle is to find out how many nuts each
had, and what were the boys' respective ages.

51.—HOW OLD WAS MARY?

Here is a funny little age problem, by the late Sam Loyd, which has been
very popular in the United States. Can you unravel the mystery?
The combined ages of Mary and Ann are forty-four years, and Mary is
twice as old as Ann was when Mary was half as old as Ann will be when
Ann is three times as old as Mary was when Mary was three times as old as
Ann. How old is Mary? That is all, but can you work it out? If not, ask your
friends to help you, and watch the shadow of bewilderment creep over their
faces as they attempt to grip the intricacies of the question.

52.—QUEER RELATIONSHIPS.
"Speaking of relationships," said the Parson at a certain dinner-party, "our
legislators are getting the marriage law into a frightful tangle, Here, for
example, is a puzzling case that has come under my notice. Two brothers
married two sisters. One man died and the other man's wife also died. Then
the survivors married."
"The man married his deceased wife's sister under the recent Act?" put in
the Lawyer.
"Exactly. And therefore, under the civil law, he is legally married and his
child is legitimate. But, you see, the man is the woman's deceased husband's
brother, and therefore, also under the civil law, she is not married to him
and her child is illegitimate."
"He is married to her and she is not married to him!" said the Doctor.
"Quite so. And the child is the legitimate son of his father, but the
illegitimate son of his mother."
"Undoubtedly 'the law is a hass,'" the Artist exclaimed, "if I may be
permitted to say so," he added, with a bow to the Lawyer.
"Certainly," was the reply. "We lawyers try our best to break in the beast to
the service of man. Our legislators are responsible for the breed."

"And this reminds me," went on the Parson, "of a man in my parish who
married the sister of his widow. This man——"
"Stop a moment, sir," said the Professor. "Married the sister of his widow?
Do you marry dead men in your parish?"
"No; but I will explain that later. Well, this man has a sister of his own.
Their names are Stephen Brown and Jane Brown. Last week a young fellow
turned up whom Stephen introduced to me as his nephew. Naturally, I spoke
of Jane as his aunt, but, to my astonishment, the youth corrected me,
assuring me that, though he was the nephew of Stephen, he was not the
nephew of Jane, the sister of Stephen. This perplexed me a good deal, but it
is quite correct."
The Lawyer was the first to get at the heart of the mystery. What was his
solution?

53.—HEARD ON THE TUBE RAILWAY.
First Lady: "And was he related to you, dear?"
Second Lady: "Oh, yes. You see, that gentleman's mother was my mother's
mother-in-law, but he is not on speaking terms with my papa."
First Lady: "Oh, indeed!" (But you could see that she was not much wiser.)
How was the gentleman related to the Second Lady?

54.—A FAMILY PARTY.
A certain family party consisted of 1 grandfather, 1 grandmother, 2 fathers,
2 mothers, 4 children, 3 grandchildren, 1 brother, 2 sisters, 2 sons, 2
daughters, 1 father-in-law, 1 mother-in-law, and 1 daughter-in-law. Twenty-
three people, you will say. No; there were only seven persons present. Can
you show how this might be?

55.—A MIXED PEDIGREE.
Joseph Bloggs: "I can't follow it, my dear boy. It makes me dizzy!"
John Snoggs: "It's very simple. Listen again! You happen to be my father's
brother-in-law, my brother's father-in-law, and also my father-in-law's
brother. You see, my father was——"
But Mr. Bloggs refused to hear any more. Can the reader show how this
extraordinary triple relationship might have come about?

56.—WILSON'S POSER.
"Speaking of perplexities——" said Mr. Wilson, throwing down a magazine
on the table in the commercial room of the Railway Hotel.
"Who was speaking of perplexities?" inquired Mr. Stubbs.
"Well, then, reading about them, if you want to be exact—it just occurred to
me that perhaps you three men may be interested in a little matter connected
with myself."
It was Christmas Eve, and the four commercial travellers were spending the
holiday at Grassminster. Probably each suspected that the others had no
homes, and perhaps each was conscious of the fact that he was in that
predicament himself. In any case they seemed to be perfectly comfortable,
and as they drew round the cheerful fire the conversation became general.
"What is the difficulty?" asked Mr. Packhurst.
"There's no difficulty in the matter, when you rightly understand it. It is like
this. A man named Parker had a flying-machine that would carry two. He
was a venturesome sort of chap—reckless, I should call him—and he had
some bother in finding a man willing to risk his life in making an ascent
with him. However, an uncle of mine thought he would chance it, and one

fine morning he took his seat in the machine and she started off well. When
they were up about a thousand feet, my nephew suddenly——"
"Here, stop, Wilson! What was your nephew doing there? You said your
uncle," interrupted Mr. Stubbs.
"Did I? Well, it does not matter. My nephew suddenly turned to Parker and
said that the engine wasn't running well, so Parker called out to my uncle
——"
"Look here," broke in Mr. Waterson, "we are getting mixed. Was it your
uncle or your nephew? Let's have it one way or the other."
"What I said is quite right. Parker called out to my uncle to do something or
other, when my nephew——"
"There you are again, Wilson," cried Mr. Stubbs; "once for all, are we to
understand that both your uncle and your nephew were on the machine?"
"Certainly. I thought I made that clear. Where was I? Well, my nephew
shouted back to Parker——"
"Phew! I'm sorry to interrupt you again, Wilson, but we can't get on like
this. Is it true that the machine would only carry two?"
"Of course. I said at the start that it only carried two."
"Then what in the name of aerostation do you mean by saying that there
were three persons on board?" shouted Mr. Stubbs.
"Who said there were three?"
"You have told us that Parker, your uncle, and your nephew went up on this
blessed flying-machine."
"That's right."
"And the thing would only carry two!"
"Right again."

"Wilson, I have known you for some time as a truthful man and a temperate
man," said Mr. Stubbs, solemnly. "But I am afraid since you took up that
new line of goods you have overworked yourself."
"Half a minute, Stubbs," interposed Mr. Waterson. "I see clearly where we
all slipped a cog. Of course, Wilson, you meant us to understand that Parker
is either your uncle or your nephew. Now we shall be all right if you will
just tell us whether Parker is your uncle or nephew."
"He is no relation to me whatever."
The three men sighed and looked anxiously at one another. Mr. Stubbs got
up from his chair to reach the matches, Mr. Packhurst proceeded to wind up
his watch, and Mr. Waterson took up the poker to attend to the fire. It was
an awkward moment, for at the season of goodwill nobody wished to tell
Mr. Wilson exactly what was in his mind.
"It's curious," said Mr. Wilson, very deliberately, "and it's rather sad, how
thick-headed some people are. You don't seem to grip the facts. It never
seems to have occurred to either of you that my uncle and my nephew are
one and the same man."
"What!" exclaimed all three together.
"Yes; David George Linklater is my uncle, and he is also my nephew.
Consequently, I am both his uncle and nephew. Queer, isn't it? I'll explain
how it comes about."
Mr. Wilson put the case so very simply that the three men saw how it might
happen without any marriage within the prohibited degrees. Perhaps the
reader can work it out for himself.

CLOCK PUZZLES.
"Look at the clock!"
Ingoldsby Legends.
In considering a few puzzles concerning clocks and watches, and the times
recorded by their hands under given conditions, it is well that a particular
convention should always be kept in mind. It is frequently the case that a
solution requires the assumption that the hands can actually record a time
involving a minute fraction of a second. Such a time, of course, cannot be
really indicated. Is the puzzle, therefore, impossible of solution? The
conclusion deduced from a logical syllogism depends for its truth on the
two premises assumed, and it is the same in mathematics. Certain things are
antecedently assumed, and the answer depends entirely on the truth of those
assumptions.
"If two horses," says Lagrange, "can pull a load of a certain weight, it is
natural to suppose that four horses could pull a load of double that weight,
six horses a load of three times that weight. Yet, strictly speaking, such is
not the case. For the inference is based on the assumption that the four
horses pull alike in amount and direction, which in practice can scarcely
ever be the case. It so happens that we are frequently led in our reckonings
to results which diverge widely from reality. But the fault is not the fault of
mathematics; for mathematics always gives back to us exactly what we
have put into it. The ratio was constant according to that supposition. The
result is founded upon that supposition. If the supposition is false the result
is necessarily false."
If one man can reap a field in six days, we say two men will reap it in three
days, and three men will do the work in two days. We here assume, as in the
case of Lagrange's horses, that all the men are exactly equally capable of
work. But we assume even more than this. For when three men get together
they may waste time in gossip or play; or, on the other hand, a spirit of
rivalry may spur them on to greater diligence. We may assume any

conditions we like in a problem, provided they be clearly expressed and
understood, and the answer will be in accordance with those conditions.

57.—WHAT WAS THE TIME?
"I say, Rackbrane, what is the time?" an acquaintance asked our friend the
professor the other day. The answer was certainly curious.
"If you add one quarter of the time from noon till now to half the time from
now till noon to-morrow, you will get the time exactly."
What was the time of day when the professor spoke?

58.—A TIME PUZZLE.
How many minutes is it until six o'clock if fifty minutes ago it was four
times as many minutes past three o'clock?

59.—A PUZZLING WATCH.
A friend pulled out his watch and said, "This watch of mine does not keep
perfect time; I must have it seen to. I have noticed that the minute hand and
the hour hand are exactly together every sixty-five minutes." Does that
watch gain or lose, and how much per hour?

60.—THE WAPSHAW'S WHARF MYSTERY.
There was a great commotion in Lower Thames Street on the morning of
January 12, 1887. When the early members of the staff arrived at

Wapshaw's Wharf they found that the safe had been broken open, a
considerable sum of money removed, and the offices left in great disorder.
The night watchman was nowhere to be found, but nobody who had been
acquainted with him for one moment suspected him to be guilty of the
robbery. In this belief the proprietors were confirmed when, later in the day,
they were informed that the poor fellow's body had been picked up by the
River Police. Certain marks of violence pointed to the fact that he had been
brutally attacked and thrown into the river. A watch found in his pocket had
stopped, as is invariably the case in such circumstances, and this was a
valuable clue to the time of the outrage. But a very stupid officer (and we
invariably find one or two stupid individuals in the most intelligent bodies
of men) had actually amused himself by turning the hands round and round,
trying to set the watch going again. After he had been severely reprimanded
for this serious indiscretion, he was asked whether he could remember the
time that was indicated by the watch when found. He replied that he could
not, but he recollected that the hour hand and minute hand were exactly
together, one above the other, and the second hand had just passed the forty-
ninth second. More than this he could not remember.
What was the exact time at which the watchman's watch stopped? The
watch is, of course, assumed to have been an accurate one.

61.—CHANGING PLACES.

The above clock face indicates a little before 42 minutes past 4. The hands
will again point at exactly the same spots a little after 23 minutes past 8. In
fact, the hands will have changed places. How many times do the hands of a
clock change places between three o'clock p.m. and midnight? And out of
all the pairs of times indicated by these changes, what is the exact time
when the minute hand will be nearest to the point IX?

62.—THE CLUB CLOCK.
One of the big clocks in the Cogitators' Club was found the other night to
have stopped just when, as will be seen in the illustration, the second hand
was exactly midway between the other two hands. One of the members
proposed to some of his friends that they should tell him the exact time
when (if the clock had not stopped) the second hand would next again have
been midway between the minute hand and the hour hand. Can you find the
correct time that it would happen?

63.—THE STOP-WATCH.

We have here a stop-watch with three hands. The second hand, which
travels once round the face in a minute, is the one with the little ring at its
end near the centre. Our dial indicates the exact time when its owner
stopped the watch. You will notice that the three hands are nearly
equidistant. The hour and minute hands point to spots that are exactly a
third of the circumference apart, but the second hand is a little too
advanced. An exact equidistance for the three hands is not possible. Now,
we want to know what the time will be when the three hands are next at
exactly the same distances as shown from one another. Can you state the
time?

64.—THE THREE CLOCKS.
On Friday, April 1, 1898, three new clocks were all set going precisely at
the same time—twelve noon. At noon on the following day it was found
that clock A had kept perfect time, that clock B had gained exactly one
minute, and that clock C had lost exactly one minute. Now, supposing that
the clocks B and C had not been regulated, but all three allowed to go on as
they had begun, and that they maintained the same rates of progress without
stopping, on what date and at what time of day would all three pairs of
hands again point at the same moment at twelve o'clock?

65.—THE RAILWAY STATION CLOCK.
A clock hangs on the wall of a railway station, 71 ft. 9 in. long and 10 ft. 4
in. high. Those are the dimensions of the wall, not of the clock! While
waiting for a train we noticed that the hands of the clock were pointing in
opposite directions, and were parallel to one of the diagonals of the wall.
What was the exact time?

66.—THE VILLAGE SIMPLETON.
A facetious individual who was taking a long walk in the country came
upon a yokel sitting on a stile. As the gentleman was not quite sure of his
road, he thought he would make inquiries of the local inhabitant; but at the
first glance he jumped too hastily to the conclusion that he had dropped on
the village idiot. He therefore decided to test the fellow's intelligence by
first putting to him the simplest question he could think of, which was,
"What day of the week is this, my good man?" The following is the smart
answer that he received:—
"When the day after to-morrow is yesterday, to-day will be as far from
Sunday as to-day was from Sunday when the day before yesterday was to-
morrow."

Can the reader say what day of the week it was? It is pretty evident that the
countryman was not such a fool as he looked. The gentleman went on his
road a puzzled but a wiser man.

LOCOMOTION AND SPEED PUZZLES.
"The race is not to the swift."—Ecclesiastes ix. II.

67.—AVERAGE SPEED.
In a recent motor ride it was found that we had gone at the rate of ten miles
an hour, but we did the return journey over the same route, owing to the
roads being more clear of traffic, at fifteen miles an hour. What was our
average speed? Do not be too hasty in your answer to this simple little
question, or it is pretty certain that you will be wrong.

68.—THE TWO TRAINS.
I put this little question to a stationmaster, and his correct answer was so
prompt that I am convinced there is no necessity to seek talented railway
officials in America or elsewhere.
Two trains start at the same time, one from London to Liverpool, the other
from Liverpool to London. If they arrive at their destinations one hour and
four hours respectively after passing one another, how much faster is one
train running than the other?

69.—THE THREE VILLAGES.
I set out the other day to ride in a motor-car from Acrefield to Butterford,
but by mistake I took the road going via Cheesebury, which is nearer
Acrefield than Butterford, and is twelve miles to the left of the direct road I

should have travelled. After arriving at Butterford I found that I had gone
thirty-five miles. What are the three distances between these villages, each
being a whole number of miles? I may mention that the three roads are
quite straight.

70.—DRAWING HER PENSION.
"Speaking of odd figures," said a gentleman who occupies some post in a
Government office, "one of the queerest characters I know is an old lame
widow who climbs up a hill every week to draw her pension at the village
post office. She crawls up at the rate of a mile and a half an hour and comes
down at the rate of four and a half miles an hour, so that it takes her just six
hours to make the double journey. Can any of you tell me how far it is from
the bottom of the hill to the top?"

71.—SIR EDWYN DE TUDOR.

In the illustration we have a sketch of Sir Edwyn de Tudor going to rescue
his lady-love, the fair Isabella, who was held a captive by a neighbouring
wicked baron. Sir Edwyn calculated that if he rode fifteen miles an hour he
would arrive at the castle an hour too soon, while if he rode ten miles an
hour he would get there just an hour too late. Now, it was of the first
importance that he should arrive at the exact time appointed, in order that
the rescue that he had planned should be a success, and the time of the tryst
was five o'clock, when the captive lady would be taking her afternoon tea.
The puzzle is to discover exactly how far Sir Edwyn de Tudor had to ride.

72.—THE HYDROPLANE QUESTION.
The inhabitants of Slocomb-on-Sea were greatly excited over the visit of a
certain flying man. All the town turned out to see the flight of the wonderful
hydroplane, and, of course, Dobson and his family were there. Master

Tommy was in good form, and informed his father that Englishmen made
better airmen than Scotsmen and Irishmen because they are not so heavy.
"How do you make that out?" asked Mr. Dobson. "Well, you see," Tommy
replied, "it is true that in Ireland there are men of Cork and in Scotland men
of Ayr, which is better still, but in England there are lightermen."
Unfortunately it had to be explained to Mrs. Dobson, and this took the edge
off the thing. The hydroplane flight was from Slocomb to the neighbouring
watering-place Poodleville—five miles distant. But there was a strong
wind, which so helped the airman that he made the outward journey in the
short time of ten minutes, though it took him an hour to get back to the
starting point at Slocomb, with the wind dead against him. Now, how long
would the ten miles have taken him if there had been a perfect calm? Of
course, the hydroplane's engine worked uniformly throughout.

73.—DONKEY RIDING.
During a visit to the seaside Tommy and Evangeline insisted on having a
donkey race over the mile course on the sands. Mr. Dobson and some of his
friends whom he had met on the beach acted as judges, but, as the donkeys
were familiar acquaintances and declined to part company the whole way, a
dead heat was unavoidable. However, the judges, being stationed at
different points on the course, which was marked off in quarter-miles, noted
the following results:—The first three-quarters were run in six and three-
quarter minutes, the first half-mile took the same time as the second half,
and the third quarter was run in exactly the same time as the last quarter.
From these results Mr. Dobson amused himself in discovering just how
long it took those two donkeys to run the whole mile. Can you give the
answer?

74.—THE BASKET OF POTATOES.

A man had a basket containing fifty potatoes. He proposed to his son, as a
little recreation, that he should place these potatoes on the ground in a
straight line. The distance between the first and second potatoes was to be
one yard, between the second and third three yards, between the third and
fourth five yards, between the fourth and fifth seven yards, and so on—an
increase of two yards for every successive potato laid down. Then the boy
was to pick them up and put them in the basket one at a time, the basket
being placed beside the first potato. How far would the boy have to travel to
accomplish the feat of picking them all up? We will not consider the
journey involved in placing the potatoes, so that he starts from the basket
with them all laid out.

75.—THE PASSENGER'S FARE.
At first sight you would hardly think there was matter for dispute in the
question involved in the following little incident, yet it took the two persons
concerned some little time to come to an agreement. Mr. Smithers hired a
motor-car to take him from Addleford to Clinkerville and back again for £3.
At Bakenham, just midway, he picked up an acquaintance, Mr. Tompkins,
and agreed to take him on to Clinkerville and bring him back to Bakenham
on the return journey. How much should he have charged the passenger?
That is the question. What was a reasonable fare for Mr. Tompkins?

DIGITAL PUZZLES.
"Nine worthies were they called."
DRYDEN: The Flower and the Leaf.
I give these puzzles, dealing with the nine digits, a class to themselves,
because I have always thought that they deserve more consideration than
they usually receive. Beyond the mere trick of "casting out nines," very
little seems to be generally known of the laws involved in these problems,
and yet an acquaintance with the properties of the digits often supplies,
among other uses, a certain number of arithmetical checks that are of real
value in the saving of labour. Let me give just one example—the first that
occurs to me.
If the reader were required to determine whether or not 15,763,530,163,289
is a square number, how would he proceed? If the number had ended with a
2, 3, 7, or 8 in the digits place, of course he would know that it could not be
a square, but there is nothing in its apparent form to prevent its being one. I
suspect that in such a case he would set to work, with a sigh or a groan, at
the laborious task of extracting the square root. Yet if he had given a little
attention to the study of the digital properties of numbers, he would settle
the question in this simple way. The sum of the digits is 59, the sum of
which is 14, the sum of which is 5 (which I call the "digital root"), and
therefore I know that the number cannot be a square, and for this reason.
The digital root of successive square numbers from 1 upwards is always 1,
4, 7, or 9, and can never be anything else. In fact, the series, 1, 4, 9, 7, 7, 9,
4, 1, 9, is repeated into infinity. The analogous series for triangular numbers
is 1, 3, 6, 1, 6, 3, 1, 9, 9. So here we have a similar negative check, for a
number cannot be triangular (that is, (n²+n)/2) if its digital root be 2, 4, 5, 7,
or 8.

76.—THE BARREL OF BEER.

A man bought an odd lot of wine in barrels and one barrel containing beer.
These are shown in the illustration, marked with the number of gallons that
each barrel contained. He sold a quantity of the wine to one man and twice
the quantity to another, but kept the beer to himself. The puzzle is to point
out which barrel contains beer. Can you say which one it is? Of course, the
man sold the barrels just as he bought them, without manipulating in any
way the contents.

77.—DIGITS AND SQUARES.

It will be seen in the diagram that we have so arranged the nine digits in a
square that the number in the second row is twice that in the first row, and
the number in the bottom row three times that in the top row. There are
three other ways of arranging the digits so as to produce the same result.
Can you find them?

78.—ODD AND EVEN DIGITS.
The odd digits, 1, 3, 5, 7, and 9, add up 25, while the even figures, 2, 4, 6,
and 8, only add up 20. Arrange these figures so that the odd ones and the
even ones add up alike. Complex and improper fractions and recurring
decimals are not allowed.

79—THE LOCKERS PUZZLE.

A man had in his office three cupboards, each containing nine lockers, as
shown in the diagram. He told his clerk to place a different one-figure
number on each locker of cupboard A, and to do the same in the case of B,
and of C. As we are here allowed to call nought a digit, and he was not
prohibited from using nought as a number, he clearly had the option of
omitting any one of ten digits from each cupboard.
Now, the employer did not say the lockers were to be numbered in any
numerical order, and he was surprised to find, when the work was done, that
the figures had apparently been mixed up indiscriminately. Calling upon his
clerk for an explanation, the eccentric lad stated that the notion had
occurred to him so to arrange the figures that in each case they formed a
simple addition sum, the two upper rows of figures producing the sum in
the lowest row. But the most surprising point was this: that he had so
arranged them that the addition in A gave the smallest possible sum, that the
addition in C gave the largest possible sum, and that all the nine digits in the
three totals were different. The puzzle is to show how this could be done.
No decimals are allowed and the nought may not appear in the hundreds
place.

80.—THE THREE GROUPS.
There appeared in "Nouvelles Annales de Mathématiques" the following
puzzle as a modification of one of my "Canterbury Puzzles." Arrange the
nine digits in three groups of two, three, and four digits, so that the first two
numbers when multiplied together make the third. Thus, 12 × 483 = 5,796. I

now also propose to include the cases where there are one, four, and four
digits, such as 4 × 1,738 = 6,952. Can you find all the possible solutions in
both cases?

81.—THE NINE COUNTERS.

I have nine counters, each bearing one of the nine digits, 1, 2, 3, 4, 5, 6, 7, 8
and 9. I arranged them on the table in two groups, as shown in the
illustration, so as to form two multiplication sums, and found that both
sums gave the same product. You will find that 158 multiplied by 23 is
3,634, and that 79 multiplied by 46 is also 3,634. Now, the puzzle I propose
is to rearrange the counters so as to get as large a product as possible. What
is the best way of placing them? Remember both groups must multiply to
the same amount, and there must be three counters multiplied by two in one
case, and two multiplied by two counters in the other, just as at present.

82.—THE TEN COUNTERS.
In this case we use the nought in addition to the 1, 2, 3, 4, 5, 6, 7, 8, 9. The
puzzle is, as in the last case, so to arrange the ten counters that the products
of the two multiplications shall be the same, and you may here have one or
more figures in the multiplier, as you choose. The above is a very easy feat;
but it is also required to find the two arrangements giving pairs of the
highest and lowest products possible. Of course every counter must be used,
and the cipher may not be placed to the left of a row of figures where it
would have no effect. Vulgar fractions or decimals are not allowed.

83.—DIGITAL MULTIPLICATION.
Here is another entertaining problem with the nine digits, the nought being
excluded. Using each figure once, and only once, we can form two
multiplication sums that have the same product, and this may be done in
many ways. For example, 7x658 and 14x329 contain all the digits once, and
the product in each case is the same—4,606. Now, it will be seen that the
sum of the digits in the product is 16, which is neither the highest nor the
lowest sum so obtainable. Can you find the solution of the problem that
gives the lowest possible sum of digits in the common product? Also that
which gives the highest possible sum?

84.—THE PIERROT'S PUZZLE.

The Pierrot in the illustration is standing in a posture that represents the sign
of multiplication. He is indicating the peculiar fact that 15 multiplied by 93
produces exactly the same figures (1,395), differently arranged. The puzzle
is to take any four digits you like (all different) and similarly arrange them
so that the number formed on one side of the Pierrot when multiplied by the
number on the other side shall produce the same figures. There are very few
ways of doing it, and I shall give all the cases possible. Can you find them
all? You are allowed to put two figures on each side of the Pierrot as in the
example shown, or to place a single figure on one side and three figures on

the other. If we only used three digits instead of four, the only possible ways
are these: 3 multiplied by 51 equals 153, and 6 multiplied by 21 equals 126.

85.—THE CAB NUMBERS.
A London policeman one night saw two cabs drive off in opposite
directions under suspicious circumstances. This officer was a particularly
careful and wide-awake man, and he took out his pocket-book to make an
entry of the numbers of the cabs, but discovered that he had lost his pencil.
Luckily, however, he found a small piece of chalk, with which he marked
the two numbers on the gateway of a wharf close by. When he returned to
the same spot on his beat he stood and looked again at the numbers, and
noticed this peculiarity, that all the nine digits (no nought) were used and
that no figure was repeated, but that if he multiplied the two numbers
together they again produced the nine digits, all once, and once only. When
one of the clerks arrived at the wharf in the early morning, he observed the
chalk marks and carefully rubbed them out. As the policeman could not
remember them, certain mathematicians were then consulted as to whether
there was any known method for discovering all the pairs of numbers that
have the peculiarity that the officer had noticed; but they knew of none. The
investigation, however, was interesting, and the following question out of
many was proposed: What two numbers, containing together all the nine
digits, will, when multiplied together, produce another number (the highest
possible) containing also all the nine digits? The nought is not allowed
anywhere.

86.—QUEER MULTIPLICATION.
If I multiply 51,249,876 by 3 (thus using all the nine digits once, and once
only), I get 153,749,628 (which again contains all the nine digits once).
Similarly, if I multiply 16,583,742 by 9 the result is 149,253,678, where in
each case all the nine digits are used. Now, take 6 as your multiplier and try

to arrange the remaining eight digits so as to produce by multiplication a
number containing all nine once, and once only. You will find it far from
easy, but it can be done.

87.—THE NUMBER CHECKS PUZZLE.
Where a large number of workmen are employed on a building it is
customary to provide every man with a little disc bearing his number. These
are hung on a board by the men as they arrive, and serve as a check on
punctuality. Now, I once noticed a foreman remove a number of these
checks from his board and place them on a split-ring which he carried in his
pocket. This at once gave me the idea for a good puzzle. In fact, I will
confide to my readers that this is just how ideas for puzzles arise. You
cannot really create an idea: it happens—and you have to be on the alert to
seize it when it does so happen.

It will be seen from the illustration that there are ten of these checks on a
ring, numbered 1 to 9 and 0. The puzzle is to divide them into three groups
without taking any off the ring, so that the first group multiplied by the
second makes the third group. For example, we can divide them into the

three groups, 2—8 9 7—1 5 4 6 3, by bringing the 6 and the 3 round to the
4, but unfortunately the first two when multiplied together do not make the
third. Can you separate them correctly? Of course you may have as many of
the checks as you like in any group. The puzzle calls for some ingenuity,
unless you have the luck to hit on the answer by chance.

88.—DIGITAL DIVISION.
It is another good puzzle so to arrange the nine digits (the nought excluded)
into two groups so that one group when divided by the other produces a
given number without remainder. For example, 1 3 4 5 8 divided by 6 7 2 9
gives 2. Can the reader find similar arrangements producing 3, 4, 5, 6, 7, 8,
and 9 respectively? Also, can he find the pairs of smallest possible numbers
in each case? Thus, 1 4 6 5 8 divided by 7 3 2 9 is just as correct for 2 as the
other example we have given, but the numbers are higher.

89.—ADDING THE DIGITS.
If I write the sum of money, £987, 5s. 4½d.., and add up the digits, they
sum to 36. No digit has thus been used a second time in the amount or
addition. This is the largest amount possible under the conditions. Now find
the smallest possible amount, pounds, shillings, pence, and farthings being
all represented. You need not use more of the nine digits than you choose,
but no digit may be repeated throughout. The nought is not allowed.

90.—THE CENTURY PUZZLE.
Can you write 100 in the form of a mixed number, using all the nine digits
once, and only once? The late distinguished French mathematician,
Edouard Lucas, found seven different ways of doing it, and expressed his

doubts as to there being any other ways. As a matter of fact there are just
eleven ways and no more. Here is one of them, 91 5742/638. Nine of the other
ways have similarly two figures in the integral part of the number, but the
eleventh expression has only one figure there. Can the reader find this last
form?

91.—MORE MIXED FRACTIONS.
When I first published my solution to the last puzzle, I was led to attempt
the expression of all numbers in turn up to 100 by a mixed fraction
containing all the nine digits. Here are twelve numbers for the reader to try
his hand at: 13, 14, 15, 16, 18, 20, 27, 36, 40, 69, 72, 94. Use every one of
the nine digits once, and only once, in every case.

92.—DIGITAL SQUARE NUMBERS.
Here are the nine digits so arranged that they form four square numbers: 9,
81, 324, 576. Now, can you put them all together so as to form a single
square number—(I) the smallest possible, and (II) the largest possible?

93.—THE MYSTIC ELEVEN.
Can you find the largest possible number containing any nine of the ten
digits (calling nought a digit) that can be divided by 11 without a
remainder? Can you also find the smallest possible number produced in the
same way that is divisible by 11? Here is an example, where the digit 5 has
been omitted: 896743012. This number contains nine of the digits and is
divisible by 11, but it is neither the largest nor the smallest number that will
work.

94.—THE DIGITAL CENTURY.
1 2 3 4 5 6 7 8 9 = 100.
It is required to place arithmetical signs between the nine figures so that
they shall equal 100. Of course, you must not alter the present numerical
arrangement of the figures. Can you give a correct solution that employs (1)
the fewest possible signs, and (2) the fewest possible separate strokes or
dots of the pen? That is, it is necessary to use as few signs as possible, and
those signs should be of the simplest form. The signs of addition and
multiplication (+ and ×) will thus count as two strokes, the sign of
subtraction (-) as one stroke, the sign of division (÷) as three, and so on.

95.—THE FOUR SEVENS.

In the illustration Professor Rackbrane is seen demonstrating one of the
little posers with which he is accustomed to entertain his class. He believes
that by taking his pupils off the beaten tracks he is the better able to secure
their attention, and to induce original and ingenious methods of thought. He
has, it will be seen, just shown how four 5's may be written with simple
arithmetical signs so as to represent 100. Every juvenile reader will see at a
glance that his example is quite correct. Now, what he wants you to do is
this: Arrange four 7's (neither more nor less) with arithmetical signs so that
they shall represent 100. If he had said we were to use four 9's we might at
once have written 999/9, but the four 7's call for rather more ingenuity. Can
you discover the little trick?

96.—THE DICE NUMBERS.

I have a set of four dice, not marked with spots in the ordinary way, but
with Arabic figures, as shown in the illustration. Each die, of course, bears
the numbers 1 to 6. When put together they will form a good many,
different numbers. As represented they make the number 1246. Now, if I
make all the different four-figure numbers that are possible with these dice
(never putting the same figure more than once in any number), what will
they all add up to? You are allowed to turn the 6 upside down, so as to
represent a 9. I do not ask, or expect, the reader to go to all the labour of
writing out the full list of numbers and then adding them up. Life is not
long enough for such wasted energy. Can you get at the answer in any other
way?

VARIOUS ARITHMETICAL AND
ALGEBRAICAL PROBLEMS.
"Variety's the very spice of life,
That gives it all its flavour."
COWPER: The Task.

97.—THE SPOT ON THE TABLE.
A boy, recently home from school, wished to give his father an exhibition of
his precocity. He pushed a large circular table into the corner of the room,
as shown in the illustration, so that it touched both walls, and he then
pointed to a spot of ink on the extreme edge.

"Here is a little puzzle for you, pater," said the youth. "That spot is exactly
eight inches from one wall and nine inches from the other. Can you tell me
the diameter of the table without measuring it?"
The boy was overheard to tell a friend, "It fairly beat the guv'nor;" but his
father is known to have remarked to a City acquaintance that he solved the
thing in his head in a minute. I often wonder which spoke the truth.

98.—ACADEMIC COURTESIES.
In a certain mixed school, where a special feature was made of the
inculcation of good manners, they had a curious rule on assembling every
morning. There were twice as many girls as boys. Every girl made a bow to
every other girl, to every boy, and to the teacher. Every boy made a bow to
every other boy, to every girl, and to the teacher. In all there were nine
hundred bows made in that model academy every morning. Now, can you
say exactly how many boys there were in the school? If you are not very
careful, you are likely to get a good deal out in your calculation.

99.—THE THIRTY-THREE PEARLS.
"A man I know," said Teddy Nicholson at a certain family party, "possesses
a string of thirty-three pearls. The middle pearl is the largest and best of all,
and the others are so selected and arranged that, starting from one end, each
successive pearl is worth £100 more than the preceding one, right up to the
big pearl. From the other end the pearls increase in value by £150 up to the
large pearl. The whole string is worth £65,000. What is the value of that
large pearl?"

"Pearls and other articles of clothing," said Uncle Walter, when the price of
the precious gem had been discovered, "remind me of Adam and Eve.
Authorities, you may not know, differ as to the number of apples that were
eaten by Adam and Eve. It is the opinion of some that Eve 8 (ate) and Adam
2 (too), a total of 10 only. But certain mathematicians have figured it out
differently, and hold that Eve 8 and Adam a total of 16. Yet the most recent

investigators think the above figures entirely wrong, for if Eve 8 and Adam
8 2, the total must be 90."
"Well," said Harry, "it seems to me that if there were giants in those days,
probably Eve 8 1 and Adam 8 2, which would give a total of 163."
"I am not at all satisfied," said Maud. "It seems to me that if Eve 8 1 and
Adam 8 1 2, they together consumed 893."
"I am sure you are all wrong," insisted Mr. Wilson, "for I consider that Eve
8 1 4 Adam, and Adam 8 1 2 4 Eve, so we get a total of 8,938."
"But, look here," broke in Herbert. "If Eve 8 1 4 Adam and Adam 8 1 2 4 2
oblige Eve, surely the total must have been 82,056!"
At this point Uncle Walter suggested that they might let the matter rest. He
declared it to be clearly what mathematicians call an indeterminate
problem.

100.—THE LABOURER'S PUZZLE.
Professor Rackbrane, during one of his rambles, chanced to come upon a
man digging a deep hole.
"Good morning," he said. "How deep is that hole?"
"Guess," replied the labourer. "My height is exactly five feet ten inches."
"How much deeper are you going?" said the professor.
"I am going twice as deep," was the answer, "and then my head will be
twice as far below ground as it is now above ground."
Rackbrane now asks if you could tell how deep that hole would be when
finished.

101.—THE TRUSSES OF HAY.
Farmer Tompkins had five trusses of hay, which he told his man Hodge to
weigh before delivering them to a customer. The stupid fellow weighed
them two at a time in all possible ways, and informed his master that the
weights in pounds were 110, 112, 113, 114, 115, 116, 117, 118, 120, and
121. Now, how was Farmer Tompkins to find out from these figures how
much every one of the five trusses weighed singly? The reader may at first
think that he ought to be told "which pair is which pair," or something of
that sort, but it is quite unnecessary. Can you give the five correct weights?

102.—MR. GUBBINS IN A FOG.
Mr. Gubbins, a diligent man of business, was much inconvenienced by a
London fog. The electric light happened to be out of order and he had to
manage as best he could with two candles. His clerk assured him that
though both were of the same length one candle would burn for four hours
and the other for five hours. After he had been working some time he put
the candles out as the fog had lifted, and he then noticed that what remained
of one candle was exactly four times the length of what was left of the
other.
When he got home that night Mr. Gubbins, who liked a good puzzle, said to
himself, "Of course it is possible to work out just how long those two
candles were burning to-day. I'll have a shot at it." But he soon found
himself in a worse fog than the atmospheric one. Could you have assisted
him in his dilemma? How long were the candles burning?

103.—PAINTING THE LAMP-POSTS.
Tim Murphy and Pat Donovan were engaged by the local authorities to
paint the lamp-posts in a certain street. Tim, who was an early riser, arrived
first on the job, and had painted three on the south side when Pat turned up

and pointed out that Tim's contract was for the north side. So Tim started
afresh on the north side and Pat continued on the south. When Pat had
finished his side he went across the street and painted six posts for Tim, and
then the job was finished. As there was an equal number of lamp-posts on
each side of the street, the simple question is: Which man painted the more
lamp-posts, and just how many more?

104.—CATCHING THE THIEF.
"Now, constable," said the defendant's counsel in cross-examination," you
say that the prisoner was exactly twenty-seven steps ahead of you when you
started to run after him?"
"Yes, sir."
"And you swear that he takes eight steps to your five?"
"That is so."
"Then I ask you, constable, as an intelligent man, to explain how you ever
caught him, if that is the case?"
"Well, you see, I have got a longer stride. In fact, two of my steps are equal
in length to five of the prisoner's. If you work it out, you will find that the
number of steps I required would bring me exactly to the spot where I
captured him."
Here the foreman of the jury asked for a few minutes to figure out the
number of steps the constable must have taken. Can you also say how many
steps the officer needed to catch the thief?

105.—THE PARISH COUNCIL ELECTION.
Here is an easy problem for the novice. At the last election of the parish
council of Tittlebury-in-the-Marsh there were twenty-three candidates for

nine seats. Each voter was qualified to vote for nine of these candidates or
for any less number. One of the electors wants to know in just how many
different ways it was possible for him to vote.

106.—THE MUDDLETOWN ELECTION.
At the last Parliamentary election at Muddletown 5,473 votes were polled.
The Liberal was elected by a majority of 18 over the Conservative, by 146
over the Independent, and by 575 over the Socialist. Can you give a simple
rule for figuring out how many votes were polled for each candidate?

107.—THE SUFFRAGISTS' MEETING.
At a recent secret meeting of Suffragists a serious difference of opinion
arose. This led to a split, and a certain number left the meeting. "I had half a
mind to go myself," said the chair-woman, "and if I had done so, two-thirds
of us would have retired." "True," said another member; "but if I had
persuaded my friends Mrs. Wild and Christine Armstrong to remain we
should only have lost half our number." Can you tell how many were
present at the meeting at the start?

108.—THE LEAP-YEAR LADIES.
Last leap-year ladies lost no time in exercising the privilege of making
proposals of marriage. If the figures that reached me from an occult source
are correct, the following represents the state of affairs in this country.
A number of women proposed once each, of whom one-eighth were
widows. In consequence, a number of men were to be married of whom
one-eleventh were widowers. Of the proposals made to widowers, one-fifth
were declined. All the widows were accepted. Thirty-five forty-fourths of

the widows married bachelors. One thousand two hundred and twenty-one
spinsters were declined by bachelors. The number of spinsters accepted by
bachelors was seven times the number of widows accepted by bachelors.
Those are all the particulars that I was able to obtain. Now, how many
women proposed?

109.—THE GREAT SCRAMBLE.
After dinner, the five boys of a household happened to find a parcel of
sugar-plums. It was quite unexpected loot, and an exciting scramble ensued,
the full details of which I will recount with accuracy, as it forms an
interesting puzzle.
You see, Andrew managed to get possession of just two-thirds of the parcel
of sugar-plums. Bob at once grabbed three-eighths of these, and Charlie
managed to seize three-tenths also. Then young David dashed upon the
scene, and captured all that Andrew had left, except one-seventh, which
Edgar artfully secured for himself by a cunning trick. Now the fun began in
real earnest, for Andrew and Charlie jointly set upon Bob, who stumbled
against the fender and dropped half of all that he had, which were equally
picked up by David and Edgar, who had crawled under a table and were
waiting. Next, Bob sprang on Charlie from a chair, and upset all the latter's
collection on to the floor. Of this prize Andrew got just a quarter, Bob
gathered up one-third, David got two-sevenths, while Charlie and Edgar
divided equally what was left of that stock.

They were just thinking the fray was over when David suddenly struck out
in two directions at once, upsetting three-quarters of what Bob and Andrew
had last acquired. The two latter, with the greatest difficulty, recovered five-
eighths of it in equal shares, but the three others each carried off one-fifth of
the same. Every sugar-plum was now accounted for, and they called a truce,
and divided equally amongst them the remainder of the parcel. What is the
smallest number of sugar-plums there could have been at the start, and what
proportion did each boy obtain?

110.—THE ABBOT'S PUZZLE.
The first English puzzlist whose name has come down to us was a
Yorkshireman—no other than Alcuin, Abbot of Canterbury (A.D. 735-804).
Here is a little puzzle from his works, which is at least interesting on

account of its antiquity. "If 100 bushels of corn were distributed among 100
people in such a manner that each man received three bushels, each woman
two, and each child half a bushel, how many men, women, and children
were there?"
Now, there are six different correct answers, if we exclude a case where
there would be no women. But let us say that there were just five times as
many women as men, then what is the correct solution?

111.—REAPING THE CORN.
A farmer had a square cornfield. The corn was all ripe for reaping, and, as
he was short of men, it was arranged that he and his son should share the
work between them. The farmer first cut one rod wide all round the square,
thus leaving a smaller square of standing corn in the middle of the field.
"Now," he said to his son, "I have cut my half of the field, and you can do
your share." The son was not quite satisfied as to the proposed division of
labour, and as the village schoolmaster happened to be passing, he appealed
to that person to decide the matter. He found the farmer was quite correct,
provided there was no dispute as to the size of the field, and on this point
they were agreed. Can you tell the area of the field, as that ingenious
schoolmaster succeeded in doing?

112.—A PUZZLING LEGACY.
A man left a hundred acres of land to be divided among his three sons—
Alfred, Benjamin, and Charles—in the proportion of one-third, one-fourth,
and one-fifth respectively. But Charles died. How was the land to be
divided fairly between Alfred and Benjamin?

113.—THE TORN NUMBER.

I had the other day in my possession a label bearing the number 3 0 2 5 in
large figures. This got accidentally torn in half, so that 3 0 was on one piece
and 2 5 on the other, as shown on the illustration. On looking at these pieces
I began to make a calculation, scarcely conscious of what I was doing,
when I discovered this little peculiarity. If we add the 3 and the 2 5 together
and square the sum we get as the result the complete original number on the
label! Thus, 30 added to 25 is 55, and 55 multiplied by 55 is 3025. Curious,
is it not? Now, the puzzle is to find another number, composed of four
figures, all different, which may be divided in the middle and produce the
same result.

114.—CURIOUS NUMBERS.
The number 48 has this peculiarity, that if you add 1 to it the result is a
square number (49, the square of 7), and if you add 1 to its half, you also
get a square number (25, the square of 5). Now, there is no limit to the
numbers that have this peculiarity, and it is an interesting puzzle to find
three more of them—the smallest possible numbers. What are they?

115.—A PRINTER'S ERROR.

In a certain article a printer had to set up the figures 54 × 23, which, of
course, means that the fourth power of 5 (625) is to be multiplied by the

cube of 2 (8), the product of which is 5,000. But he printed 54 × 23 as 5 4 2
3, which is not correct. Can you place four digits in the manner shown, so
that it will be equally correct if the printer sets it up aright or makes the
same blunder?

116.—THE CONVERTED MISER.
Mr. Jasper Bullyon was one of the very few misers who have ever been
converted to a sense of their duty towards their less fortunate fellow-men.
One eventful night he counted out his accumulated wealth, and resolved to
distribute it amongst the deserving poor.
He found that if he gave away the same number of pounds every day in the
year, he could exactly spread it over a twelvemonth without there being
anything left over; but if he rested on the Sundays, and only gave away a
fixed number of pounds every weekday, there would be one sovereign left
over on New Year's Eve. Now, putting it at the lowest possible, what was
the exact number of pounds that he had to distribute?
Could any question be simpler? A sum of pounds divided by one number of
days leaves no remainder, but divided by another number of days leaves a
sovereign over. That is all; and yet, when you come to tackle this little
question, you will be surprised that it can become so puzzling.

117.—A FENCE PROBLEM.
The practical usefulness of puzzles is a point that we are liable to overlook.
Yet, as a matter of fact, I have from time to time received quite a large
number of letters from individuals who have found that the mastering of
some little principle upon which a puzzle was built has proved of
considerable value to them in a most unexpected way. Indeed, it may be
accepted as a good maxim that a puzzle is of little real value unless, as well
as being amusing and perplexing, it conceals some instructive and possibly

useful feature. It is, however, very curious how these little bits of acquired
knowledge dovetail into the occasional requirements of everyday life, and
equally curious to what strange and mysterious uses some of our readers
seem to apply them. What, for example, can be the object of Mr. Wm.
Oxley, who writes to me all the way from Iowa, in wishing to ascertain the
dimensions of a field that he proposes to enclose, containing just as many
acres as there shall be rails in the fence?

The man wishes to fence in a perfectly square field which is to contain just
as many acres as there are rails in the required fence. Each hurdle, or
portion of fence, is seven rails high, and two lengths would extend one pole
(16½ ft.): that is to say, there are fourteen rails to the pole, lineal measure.
Now, what must be the size of the field?

118.—CIRCLING THE SQUARES.

The puzzle is to place a different number in each of the ten squares so that
the sum of the squares of any two adjacent numbers shall be equal to the
sum of the squares of the two numbers diametrically opposite to them. The
four numbers placed, as examples, must stand as they are. The square of 16
is 256, and the square of 2 is 4. Add these together, and the result is 260.
Also—the square of 14 is 196, and the square of 8 is 64. These together also
make 260. Now, in precisely the same way, B and C should be equal to G
and H (the sum will not necessarily be 260), A and K to F and E, H and I to
C and D, and so on, with any two adjoining squares in the circle.
All you have to do is to fill in the remaining six numbers. Fractions are not
allowed, and I shall show that no number need contain more than two
figures.

119.—RACKBRANE'S LITTLE LOSS.

Professor Rackbrane was spending an evening with his old friends, Mr. and
Mrs. Potts, and they engaged in some game (he does not say what game) of
cards. The professor lost the first game, which resulted in doubling the
money that both Mr. and Mrs. Potts had laid on the table. The second game
was lost by Mrs. Potts, which doubled the money then held by her husband
and the professor. Curiously enough, the third game was lost by Mr. Potts,
and had the effect of doubling the money then held by his wife and the
professor. It was then found that each person had exactly the same money,
but the professor had lost five shillings in the course of play. Now, the
professor asks, what was the sum of money with which he sat down at the
table? Can you tell him?

120.—THE FARMER AND HIS SHEEP.

Farmer Longmore had a curious aptitude for arithmetic, and was known in
his district as the "mathematical farmer." The new vicar was not aware of
this fact when, meeting his worthy parishioner one day in the lane, he asked

him in the course of a short conversation, "Now, how many sheep have you
altogether?" He was therefore rather surprised at Longmore's answer, which
was as follows: "You can divide my sheep into two different parts, so that
the difference between the two numbers is the same as the difference
between their squares. Maybe, Mr. Parson, you will like to work out the
little sum for yourself."
Can the reader say just how many sheep the farmer had? Supposing he had
possessed only twenty sheep, and he divided them into the two parts 12 and
8. Now, the difference between their squares, 144 and 64, is 80. So that will
not do, for 4 and 80 are certainly not the same. If you can find numbers that
work out correctly, you will know exactly how many sheep Farmer
Longmore owned.

121.—HEADS OR TAILS.
Crooks, an inveterate gambler, at Goodwood recently said to a friend, "I'll
bet you half the money in my pocket on the toss of a coin—heads I win,
tails I lose." The coin was tossed and the money handed over. He repeated
the offer again and again, each time betting half the money then in his
possession. We are not told how long the game went on, or how many times
the coin was tossed, but this we know, that the number of times that Crooks
lost was exactly equal to the number of times that he won. Now, did he gain
or lose by this little venture?

122.—THE SEE-SAW PUZZLE.
Necessity is, indeed, the mother of invention. I was amused the other day in
watching a boy who wanted to play see-saw and, in his failure to find
another child to share the sport with him, had been driven back upon the
ingenious resort of tying a number of bricks to one end of the plank to
balance his weight at the other.

As a matter of fact, he just balanced against sixteen bricks, when these were
fixed to the short end of plank, but if he fixed them to the long end of plank
he only needed eleven as balance.
Now, what was that boy's weight, if a brick weighs equal to a three-quarter
brick and three-quarters of a pound?

123.—A LEGAL DIFFICULTY.
"A client of mine," said a lawyer, "was on the point of death when his wife
was about to present him with a child. I drew up his will, in which he
settled two-thirds of his estate upon his son (if it should happen to be a boy)
and one-third on the mother. But if the child should be a girl, then two-
thirds of the estate should go to the mother and one-third to the daughter. As
a matter of fact, after his death twins were born—a boy and a girl. A very
nice point then arose. How was the estate to be equitably divided among the
three in the closest possible accordance with the spirit of the dead man's
will?"

124.—A QUESTION OF DEFINITION.
"My property is exactly a mile square," said one landowner to another.
"Curiously enough, mine is a square mile," was the reply.
"Then there is no difference?"
Is this last statement correct?

125.—THE MINERS' HOLIDAY.

Seven coal-miners took a holiday at the seaside during a big strike. Six of
the party spent exactly half a sovereign each, but Bill Harris was more
extravagant. Bill spent three shillings more than the average of the party.
What was the actual amount of Bill's expenditure?

126.—SIMPLE MULTIPLICATION.
If we number six cards 1, 2, 4, 5, 7, and 8, and arrange them on the table in
this order:—
1 4 2 8 5 7
We can demonstrate that in order to multiply by 3 all that is necessary is to
remove the 1 to the other end of the row, and the thing is done. The answer
is 428571. Can you find a number that, when multiplied by 3 and divided
by 2, the answer will be the same as if we removed the first card (which in
this case is to be a 3) From the beginning of the row to the end?

127.—SIMPLE DIVISION.
Sometimes a very simple question in elementary arithmetic will cause a
good deal of perplexity. For example, I want to divide the four numbers,
701, 1,059, 1,417, and 2,312, by the largest number possible that will leave
the same remainder in every case. How am I to set to work Of course, by a
laborious system of trial one can in time discover the answer, but there is
quite a simple method of doing it if you can only find it.

128.—A PROBLEM IN SQUARES.
We possess three square boards. The surface of the first contains five square
feet more than the second, and the second contains five square feet more

than the third. Can you give exact measurements for the sides of the
boards? If you can solve this little puzzle, then try to find three squares in
arithmetical progression, with a common difference of 7 and also of 13.

129.—THE BATTLE OF HASTINGS.
All historians know that there is a great deal of mystery and uncertainty
concerning the details of the ever-memorable battle on that fatal day,
October 14, 1066. My puzzle deals with a curious passage in an ancient
monkish chronicle that may never receive the attention that it deserves, and
if I am unable to vouch for the authenticity of the document it will none the
less serve to furnish us with a problem that can hardly fail to interest those
of my readers who have arithmetical predilections. Here is the passage in
question.
"The men of Harold stood well together, as their wont was, and formed
sixty and one squares, with a like number of men in every square thereof,
and woe to the hardy Norman who ventured to enter their redoubts; for a
single blow of a Saxon war-hatchet would break his lance and cut through
his coat of mail.... When Harold threw himself into the fray the Saxons
were one mighty square of men, shouting the battle-cries, 'Ut!' 'Olicrosse!'
'Godemitè!'"
Now, I find that all the contemporary authorities agree that the Saxons did
actually fight in this solid order. For example, in the "Carmen de Bello
Hastingensi," a poem attributed to Guy, Bishop of Amiens, living at the
time of the battle, we are told that "the Saxons stood fixed in a dense mass,"
and Henry of Huntingdon records that "they were like unto a castle,
impenetrable to the Normans;" while Robert Wace, a century after, tells us
the same thing. So in this respect my newly-discovered chronicle may not
be greatly in error. But I have reason to believe that there is something
wrong with the actual figures. Let the reader see what he can make of them.
The number of men would be sixty-one times a square number; but when
Harold himself joined in the fray they were then able to form one large
square. What is the smallest possible number of men there could have been?

In order to make clear to the reader the simplicity of the question, I will
give the lowest solutions in the case of 60 and 62, the numbers immediately
preceding and following 61. They are 60 × 42 + 1 = 312, and
62 × 82 + 1 = 632. That is, 60 squares of 16 men each would be 960 men,
and when Harold joined them they would be 961 in number, and so form a
square with 31 men on every side. Similarly in the case of the figures I have
given for 62. Now, find the lowest answer for 61.

130.—THE SCULPTOR'S PROBLEM.
An ancient sculptor was commissioned to supply two statues, each on a
cubical pedestal. It is with these pedestals that we are concerned. They were
of unequal sizes, as will be seen in the illustration, and when the time
arrived for payment a dispute arose as to whether the agreement was based
on lineal or cubical measurement. But as soon as they came to measure the
two pedestals the matter was at once settled, because, curiously enough, the
number of lineal feet was exactly the same as the number of cubical feet.
The puzzle is to find the dimensions for two pedestals having this
peculiarity, in the smallest possible figures. You see, if the two pedestals,
for example, measure respectively 3 ft. and 1 ft. on every side, then the
lineal measurement would be 4 ft. and the cubical contents 28 ft., which are
not the same, so these measurements will not do.

131.—THE SPANISH MISER.
There once lived in a small town in New Castile a noted miser named Don
Manuel Rodriguez. His love of money was only equalled by a strong
passion for arithmetical problems. These puzzles usually dealt in some way
or other with his accumulated treasure, and were propounded by him solely
in order that he might have the pleasure of solving them himself.
Unfortunately very few of them have survived, and when travelling through
Spain, collecting material for a proposed work on "The Spanish Onion as a
Cause of National Decadence," I only discovered a very few. One of these
concerns the three boxes that appear in the accompanying authentic portrait.

Each box contained a different number of golden doubloons. The difference
between the number of doubloons in the upper box and the number in the
middle box was the same as the difference between the number in the
middle box and the number in the bottom box. And if the contents of any
two of the boxes were united they would form a square number. What is the
smallest number of doubloons that there could have been in any one of the
boxes?

132.—THE NINE TREASURE BOXES.
The following puzzle will illustrate the importance on occasions of being
able to fix the minimum and maximum limits of a required number. This
can very frequently be done. For example, it has not yet been ascertained in

how many different ways the knight's tour can be performed on the chess
board; but we know that it is fewer than the number of combinations of 168
things taken 63 at a time and is greater than 31,054,144—for the latter is the
number of routes of a particular type. Or, to take a more familiar case, if
you ask a man how many coins he has in his pocket, he may tell you that he
has not the slightest idea. But on further questioning you will get out of him
some such statement as the following: "Yes, I am positive that I have more
than three coins, and equally certain that there are not so many as twenty-
five." Now, the knowledge that a certain number lies between 2 and 12 in
my puzzle will enable the solver to find the exact answer; without that
information there would be an infinite number of answers, from which it
would be impossible to select the correct one.
This is another puzzle received from my friend Don Manuel Rodriguez, the
cranky miser of New Castile. On New Year's Eve in 1879 he showed me
nine treasure boxes, and after informing me that every box contained a
square number of golden doubloons, and that the difference between the
contents of A and B was the same as between B and C, D and E, E and F, G
and H, or H and I, he requested me to tell him the number of coins in every
one of the boxes. At first I thought this was impossible, as there would be
an infinite number of different answers, but on consideration I found that
this was not the case. I discovered that while every box contained coins, the
contents of A, B, C increased in weight in alphabetical order; so did D, E, F;
and so did G, H, I; but D or E need not be heavier than C, nor G or H
heavier than F. It was also perfectly certain that box A could not contain
more than a dozen coins at the outside; there might not be half that number,
but I was positive that there were not more than twelve. With this
knowledge I was able to arrive at the correct answer.
In short, we have to discover nine square numbers such that A, B, C; and D,
E, F; and G, H, I are three groups in arithmetical progression, the common
difference being the same in each group, and A being less than 12. How
many doubloons were there in every one of the nine boxes?

133.—THE FIVE BRIGANDS.

The five Spanish brigands, Alfonso, Benito, Carlos, Diego, and Esteban,
were counting their spoils after a raid, when it was found that they had
captured altogether exactly 200 doubloons. One of the band pointed out that
if Alfonso had twelve times as much, Benito three times as much, Carlos
the same amount, Diego half as much, and Esteban one-third as much, they
would still have altogether just 200 doubloons. How many doubloons had
each?
There are a good many equally correct answers to this question. Here is one
of them:

A 6× 12= 72
B 12× 3= 36
C 17× 1= 17
D 120× ½= 60
E 45 × 1/3 = 15
200 200

The puzzle is to discover exactly how many different answers there are, it
being understood that every man had something and that there is to be no
fractional money—only doubloons in every case.
This problem, worded somewhat differently, was propounded by Tartaglia
(died 1559), and he flattered himself that he had found one solution; but a
French mathematician of note (M.A. Labosne), in a recent work, says that
his readers will be astonished when he assures them that there are 6,639
different correct answers to the question. Is this so? How many answers are
there?

134.—THE BANKER'S PUZZLE.
A banker had a sporting customer who was always anxious to wager on
anything. Hoping to cure him of his bad habit, he proposed as a wager that
the customer would not be able to divide up the contents of a box
containing only sixpences into an exact number of equal piles of sixpences.

The banker was first to put in one or more sixpences (as many as he liked);
then the customer was to put in one or more (but in his case not more than a
pound in value), neither knowing what the other put in. Lastly, the customer
was to transfer from the banker's counter to the box as many sixpences as
the banker desired him to put in. The puzzle is to find how many sixpences
the banker should first put in and how many he should ask the customer to
transfer, so that he may have the best chance of winning.

135.—THE STONEMASON'S PROBLEM.
A stonemason once had a large number of cubic blocks of stone in his yard,
all of exactly the same size. He had some very fanciful little ways, and one
of his queer notions was to keep these blocks piled in cubical heaps, no two
heaps containing the same number of blocks. He had discovered for himself
(a fact that is well known to mathematicians) that if he took all the blocks
contained in any number of heaps in regular order, beginning with the
single cube, he could always arrange those on the ground so as to form a
perfect square. This will be clear to the reader, because one block is a
square, 1 + 8 = 9 is a square, 1 + 8 + 27 = 36 is a square,
1 + 8 + 27 + 64 = 100 is a square, and so on. In fact, the sum of any number
of consecutive cubes, beginning always with 1, is in every case a square
number.
One day a gentleman entered the mason's yard and offered him a certain
price if he would supply him with a consecutive number of these cubical
heaps which should contain altogether a number of blocks that could be laid
out to form a square, but the buyer insisted on more than three heaps and
declined to take the single block because it contained a flaw. What was the
smallest possible number of blocks of stone that the mason had to supply?

136.—THE SULTAN'S ARMY.

A certain Sultan wished to send into battle an army that could be formed
into two perfect squares in twelve different ways. What is the smallest
number of men of which that army could be composed? To make it clear to
the novice, I will explain that if there were 130 men, they could be formed
into two squares in only two different ways—81 and 49, or 121 and 9. Of
course, all the men must be used on every occasion.

137.—A STUDY IN THRIFT.
Certain numbers are called triangular, because if they are taken to represent
counters or coins they may be laid out on the table so as to form triangles.
The number 1 is always regarded as triangular, just as 1 is a square and a
cube number. Place one counter on the table—that is, the first triangular
number. Now place two more counters beneath it, and you have a triangle
of three counters; therefore 3 is triangular. Next place a row of three more
counters, and you have a triangle of six counters; therefore 6 is triangular.
We see that every row of counters that we add, containing just one more
counter than the row above it, makes a larger triangle.
Now, half the sum of any number and its square is always a triangular
number. Thus half of 2 + 22 = 3; half of 3 + 32 = 6; half of 4 + 42 = 10; half
of 5 + 52= 15; and so on. So if we want to form a triangle with 8 counters
on each side we shall require half of 8 + 82, or 36 counters. This is a pretty
little property of numbers. Before going further, I will here say that if the
reader refers to the "Stonemason's Problem" (No. 135) he will remember
that the sum of any number of consecutive cubes beginning with 1 is always
a square, and these form the series 12, 32, 62, 102, etc. It will now be
understood when I say that one of the keys to the puzzle was the fact that
these are always the squares of triangular numbers—that is, the squares of
1, 3, 6, 10, 15, 21, 28, etc., any of which numbers we have seen will form a
triangle.
Every whole number is either triangular, or the sum of two triangular
numbers or the sum of three triangular numbers. That is, if we take any
number we choose we can always form one, two, or three triangles with

them. The number 1 will obviously, and uniquely, only form one triangle;
some numbers will only form two triangles (as 2, 4, 11, etc.); some numbers
will only form three triangles (as 5, 8, 14, etc.). Then, again, some numbers
will form both one and two triangles (as 6), others both one and three
triangles (as 3 and 10), others both two and three triangles (as 7 and 9),
while some numbers (like 21) will form one, two, or three triangles, as we
desire. Now for a little puzzle in triangular numbers.
Sandy McAllister, of Aberdeen, practised strict domestic economy, and was
anxious to train his good wife in his own habits of thrift. He told her last
New Year's Eve that when she had saved so many sovereigns that she could
lay them all out on the table so as to form a perfect square, or a perfect
triangle, or two triangles, or three triangles, just as he might choose to ask
he would add five pounds to her treasure. Soon she went to her husband
with a little bag of £36 in sovereigns and claimed her reward. It will be
found that the thirty-six coins will form a square (with side 6), that they will
form a single triangle (with side 8), that they will form two triangles (with
sides 5 and 6), and that they will form three triangles (with sides 3, 5, and
5). In each of the four cases all the thirty-six coins are used, as required, and
Sandy therefore made his wife the promised present like an honest man.
The Scotsman then undertook to extend his promise for five more years, so
that if next year the increased number of sovereigns that she has saved can
be laid out in the same four different ways she will receive a second
present; if she succeeds in the following year she will get a third present,
and so on until she has earned six presents in all. Now, how many
sovereigns must she put together before she can win the sixth present?
What you have to do is to find five numbers, the smallest possible, higher
than 36, that can be displayed in the four ways—to form a square, to form a
triangle, to form two triangles, and to form three triangles. The highest of
your five numbers will be your answer.

138.—THE ARTILLERYMEN'S DILEMMA.

"All cannon-balls are to be piled in square pyramids," was the order issued
to the regiment. This was done. Then came the further order, "All pyramids
are to contain a square number of balls." Whereupon the trouble arose. "It
can't be done," said the major. "Look at this pyramid, for example; there are
sixteen balls at the base, then nine, then four, then one at the top, making
thirty balls in all. But there must be six more balls, or five fewer, to make a
square number." "It must be done," insisted the general. "All you have to do
is to put the right number of balls in your pyramids." "I've got it!" said a
lieutenant, the mathematical genius of the regiment. "Lay the balls out
singly." "Bosh!" exclaimed the general. "You can't pile one ball into a
pyramid!" Is it really possible to obey both orders?

139.—THE DUTCHMEN'S WIVES.
I wonder how many of my readers are acquainted with the puzzle of the
"Dutchmen's Wives"—in which you have to determine the names of three
men's wives, or, rather, which wife belongs to each husband. Some thirty
years ago it was "going the rounds," as something quite new, but I recently
discovered it in the Ladies' Diary for 1739-40, so it was clearly familiar to

the fair sex over one hundred and seventy years ago. How many of our
mothers, wives, sisters, daughters, and aunts could solve the puzzle to-day?
A far greater proportion than then, let us hope.
Three Dutchmen, named Hendrick, Elas, and Cornelius, and their wives,
Gurtrün, Katrün, and Anna, purchase hogs. Each buys as many as he (or
she) gives shillings for one. Each husband pays altogether three guineas
more than his wife. Hendrick buys twenty-three more hogs than Katrün, and
Elas eleven more than Gurtrün. Now, what was the name of each man's
wife?

140.—FIND ADA'S SURNAME.
This puzzle closely resembles the last one, my remarks on the solution of
which the reader may like to apply in another case. It was recently
submitted to a Sydney evening newspaper that indulges in "intellect
sharpeners," but was rejected with the remark that it is childish and that
they only published problems capable of solution! Five ladies, accompanied

by their daughters, bought cloth at the same shop. Each of the ten paid as
many farthings per foot as she bought feet, and each mother spent 8s. 5¼d.
more than her daughter. Mrs. Robinson spent 6s. more than Mrs. Evans,
who spent about a quarter as much as Mrs. Jones. Mrs. Smith spent most of
all. Mrs. Brown bought 21 yards more than Bessie—one of the girls. Annie
bought 16 yards more than Mary and spent £3, 0s. 8d. more than Emily. The
Christian name of the other girl was Ada. Now, what was her surname?

141.—SATURDAY MARKETING.
Here is an amusing little case of marketing which, although it deals with a
good many items of money, leads up to a question of a totally different
character. Four married couples went into their village on a recent Saturday
night to do a little marketing. They had to be very economical, for among
them they only possessed forty shilling coins. The fact is, Ann spent 1s.,
Mary spent 2s., Jane spent 3s., and Kate spent 4s. The men were rather
more extravagant than their wives, for Ned Smith spent as much as his wife,
Tom Brown twice as much as his wife, Bill Jones three times as much as his
wife, and Jack Robinson four times as much as his wife. On the way home
somebody suggested that they should divide what coin they had left equally
among them. This was done, and the puzzling question is simply this: What
was the surname of each woman? Can you pair off the four couples?

GEOMETRICAL PROBLEMS.
"God geometrizes continually."
PLATO.
"There is no study," said Augustus de Morgan, "which presents so simple a
beginning as that of geometry; there is none in which difficulties grow more
rapidly as we proceed." This will be found when the reader comes to
consider the following puzzles, though they are not arranged in strict order
of difficulty. And the fact that they have interested and given pleasure to
man for untold ages is no doubt due in some measure to the appeal they
make to the eye as well as to the brain. Sometimes an algebraical formula or
theorem seems to give pleasure to the mathematician's eye, but it is
probably only an intellectual pleasure. But there can be no doubt that in the
case of certain geometrical problems, notably dissection or superposition
puzzles, the æsthetic faculty in man contributes to the delight. For example,
there are probably few readers who will examine the various cuttings of the
Greek cross in the following pages without being in some degree stirred by
a sense of beauty. Law and order in Nature are always pleasing to
contemplate, but when they come under the very eye they seem to make a
specially strong appeal. Even the person with no geometrical knowledge
whatever is induced after the inspection of such things to exclaim, "How
very pretty!" In fact, I have known more than one person led on to a study
of geometry by the fascination of cutting-out puzzles. I have, therefore,
thought it well to keep these dissection puzzles distinct from the
geometrical problems on more general lines.

DISSECTION PUZZLES.
"Take him and cut him out in little stars."
Romeo and Juliet, iii. 2.
Puzzles have infinite variety, but perhaps there is no class more ancient than
dissection, cutting-out, or superposition puzzles. They were certainly
known to the Chinese several thousand years before the Christian era. And
they are just as fascinating to-day as they can have been at any period of
their history. It is supposed by those who have investigated the matter that
the ancient Chinese philosophers used these puzzles as a sort of
kindergarten method of imparting the principles of geometry. Whether this
was so or not, it is certain that all good dissection puzzles (for the nursery
type of jig-saw puzzle, which merely consists in cutting up a picture into
pieces to be put together again, is not worthy of serious consideration) are
really based on geometrical laws. This statement need not, however,
frighten off the novice, for it means little more than this, that geometry will
give us the "reason why," if we are interested in knowing it, though the
solutions may often be discovered by any intelligent person after the
exercise of patience, ingenuity, and common sagacity.
If we want to cut one plane figure into parts that by readjustment will form
another figure, the first thing is to find a way of doing it at all, and then to
discover how to do it in the fewest possible pieces. Often a dissection
problem is quite easy apart from this limitation of pieces. At the time of the
publication in the Weekly Dispatch, in 1902, of a method of cutting an
equilateral triangle into four parts that will form a square (see No. 26,
"Canterbury Puzzles"), no geometrician would have had any difficulty in
doing what is required in five pieces: the whole point of the discovery lay in
performing the little feat in four pieces only.
Mere approximations in the case of these problems are valueless; the
solution must be geometrically exact, or it is not a solution at all. Fallacies
are cropping up now and again, and I shall have occasion to refer to one or
two of these. They are interesting merely as fallacies. But I want to say

something on two little points that are always arising in cutting-out puzzles
—the questions of "hanging by a thread" and "turning over." These points
can best be illustrated by a puzzle that is frequently to be found in the old
books, but invariably with a false solution. The puzzle is to cut the figure
shown in Fig. 1 into three pieces that will fit together and form a half-
square triangle. The answer that is invariably given is that shown in Figs. 1
and 2. Now, it is claimed that the four pieces marked C are really only one
piece, because they may be so cut that they are left "hanging together by a
mere thread." But no serious puzzle lover will ever admit this. If the cut is
made so as to leave the four pieces joined in one, then it cannot result in a
perfectly exact solution. If, on the other hand, the solution is to be exact,
then there will be four pieces—or six pieces in all. It is, therefore, not a
solution in three pieces.

If, however, the reader will look at the solution in Figs. 3 and 4, he will see
that no such fault can be found with it. There is no question whatever that
there are three pieces, and the solution is in this respect quite satisfactory.

But another question arises. It will be found on inspection that the piece
marked F, in Fig. 3, is turned over in Fig. 4—that is to say, a different side
has necessarily to be presented. If the puzzle were merely to be cut out of
cardboard or wood, there might be no objection to this reversal, but it is
quite possible that the material would not admit of being reversed. There
might be a pattern, a polish, a difference of texture, that prevents it. But it is
generally understood that in dissection puzzles you are allowed to turn
pieces over unless it is distinctly stated that you may not do so. And very
often a puzzle is greatly improved by the added condition, "no piece may be
turned over." I have often made puzzles, too, in which the diagram has a
small repeated pattern, and the pieces have then so to be cut that not only is
there no turning over, but the pattern has to be matched, which cannot be
done if the pieces are turned round, even with the proper side uppermost.
Before presenting a varied series of cutting-out puzzles, some very easy and
others difficult, I propose to consider one family alone—those problems
involving what is known as the Greek cross with the square. This will
exhibit a great variety of curious transpositions, and, by having the
solutions as we go along, the reader will be saved the trouble of perpetually
turning to another part of the book, and will have everything under his eye.
It is hoped that in this way the article may prove somewhat instructive to
the novice and interesting to others.

GREEK CROSS PUZZLES.
"To fret thy soul with crosses."
SPENSER.
"But, for my part, it was Greek to me."
Julius Cæsar, i. 2.
Many people are accustomed to consider the cross as a wholly Christian
symbol. This is erroneous: it is of very great antiquity. The ancient
Egyptians employed it as a sacred symbol, and on Greek sculptures we find
representations of a cake (the supposed real origin of our hot cross buns)
bearing a cross. Two such cakes were discovered at Herculaneum. Cecrops
offered to Jupiter Olympus a sacred cake or boun of this kind. The cross
and ball, so frequently found on Egyptian figures, is a circle and the tau
cross. The circle signified the eternal preserver of the world, and the T,
named from the Greek letter tau, is the monogram of Thoth, the Egyptian
Mercury, meaning wisdom. This tau cross is also called by Christians the
cross of St. Anthony, and is borne on a badge in the bishop's palace at
Exeter. As for the Greek or mundane cross, the cross with four equal arms,
we are told by competent antiquaries that it was regarded by ancient
occultists for thousands of years as a sign of the dual forces of Nature—the
male and female spirit of everything that was everlasting.

The Greek cross, as shown in Fig. 5, is formed by the assembling together
of five equal squares. We will start with what is known as the Hindu
problem, supposed to be upwards of three thousand years old. It appears in
the seal of Harvard College, and is often given in old works as symbolical
of mathematical science and exactitude. Cut the cross into five pieces to
form a square. Figs. 6 and 7 show how this is done. It was not until the
middle of the nineteenth century that we found that the cross might be
transformed into a square in only four pieces. Figs. 8 and 9 will show how
to do it, if we further require the four pieces to be all of the same size and
shape. This Fig. 9 is remarkable because, according to Dr. Le Plongeon and

others, as expounded in a work by Professor Wilson of the Smithsonian
Institute, here we have the great Swastika, or sign, of "good luck to you "—
the most ancient symbol of the human race of which there is any record.
Professor Wilson's work gives some four hundred illustrations of this
curious sign as found in the Aztec mounds of Mexico, the pyramids of
Egypt, the ruins of Troy, and the ancient lore of India and China. One might
almost say there is a curious affinity between the Greek cross and Swastika!
If, however, we require that the four pieces shall be produced by only two
clips of the scissors (assuming the puzzle is in paper form), then we must
cut as in Fig. 10 to form Fig. 11, the first clip of the scissors being from a to
b. Of course folding the paper, or holding the pieces together after the first
cut, would not in this case be allowed. But there is an infinite number of
different ways of making the cuts to solve the puzzle in four pieces. To this
point I propose to return.

It will be seen that every one of these puzzles has its reverse puzzle—to cut
a square into pieces to form a Greek cross. But as a square has not so many
angles as the cross, it is not always equally easy to discover the true
directions of the cuts. Yet in the case of the examples given, I will leave the
reader to determine their direction for himself, as they are rather obvious
from the diagrams.

Cut a square into five pieces that will form two separate Greek crosses of
different sizes. This is quite an easy puzzle. As will be seen in Fig. 12, we
have only to divide our square into 25 little squares and then cut as shown.
The cross A is cut out entire, and the pieces B, C, D, and E form the larger
cross in Fig. 13. The reader may here like to cut the single piece, B, into
four pieces all similar in shape to itself, and form a cross with them in the
manner shown in Fig. 13. I hardly need give the solution.

Cut a square into five pieces that will form two separate Greek crosses of
exactly the same size. This is more difficult. We make the cuts as in Fig. 14,
where the cross A comes out entire and the other four pieces form the cross

in Fig. 15. The direction of the cuts is pretty obvious. It will be seen that the
sides of the square in Fig. 14 are marked off into six equal parts. The sides
of the cross are found by ruling lines from certain of these points to others.
I will now explain, as I promised, why a Greek cross may be cut into four
pieces in an infinite number of different ways to make a square. Draw a
cross, as in Fig. 16. Then draw on transparent paper the square shown in
Fig. 17, taking care that the distance c to d is exactly the same as the
distance a to b in the cross. Now place the transparent paper over the cross
and slide it about into different positions, only be very careful always to
keep the square at the same angle to the cross as shown, where a b is
parallel to c d. If you place the point c exactly over a the lines will indicate
the solution (Figs. 10 and 11). If you place c in the very centre of the dotted
square, it will give the solution in Figs. 8 and 9. You will now see that by
sliding the square about so that the point c is always within the dotted
square you may get as many different solutions as you like; because, since
an infinite number of different points may theoretically be placed within
this square, there must be an infinite number of different solutions. But the
point c need not necessarily be placed within the dotted square. It may be
placed, for example, at point e to give a solution in four pieces. Here the
joins at a and f may be as slender as you like. Yet if you once get over the
edge at a or f you no longer have a solution in four pieces. This proof will
be found both entertaining and instructive. If you do not happen to have any
transparent paper at hand, any thin paper will of course do if you hold the
two sheets against a pane of glass in the window.

It may have been noticed from the solutions of the puzzles that I have given
that the side of the square formed from the cross is always equal to the
distance a to b in Fig. 16. This must necessarily be so, and I will presently
try to make the point quite clear.
We will now go one step further. I have already said that the ideal solution
to a cutting-out puzzle is always that which requires the fewest possible
pieces. We have just seen that two crosses of the same size may be cut out
of a square in five pieces. The reader who succeeded in solving this perhaps
asked himself: "Can it be done in fewer pieces?" This is just the sort of
question that the true puzzle lover is always asking, and it is the right
attitude for him to adopt. The answer to the question is that the puzzle may
be solved in four pieces—the fewest possible. This, then, is a new puzzle.
Cut a square into four pieces that will form two Greek crosses of the same
size.

The solution is very beautiful. If you divide by points the sides of the square
into three equal parts, the directions of the lines in Fig. 18 will be quite
obvious. If you cut along these lines, the pieces A and B will form the cross
in Fig. 19 and the pieces C and D the similar cross in Fig. 20. In this square
we have another form of Swastika.
The reader will here appreciate the truth of my remark to the effect that it is
easier to find the directions of the cuts when transforming a cross to a
square than when converting a square into a cross. Thus, in Figs. 6, 8, and
10 the directions of the cuts are more obvious than in Fig. 14, where we had
first to divide the sides of the square into six equal parts, and in Fig. 18,
where we divide them into three equal parts. Then, supposing you were
required to cut two equal Greek crosses, each into two pieces, to form a
square, a glance at Figs. 19 and 20 will show how absurdly more easy this
is than the reverse puzzle of cutting the square to make two crosses.
Referring to my remarks on "fallacies," I will now give a little example of
these "solutions" that are not solutions. Some years ago a young
correspondent sent me what he evidently thought was a brilliant new
discovery—the transforming of a square into a Greek cross in four pieces
by cuts all parallel to the sides of the square. I give his attempt in Figs. 21
and 22, where it will be seen that the four pieces do not form a symmetrical
Greek cross, because the four arms are not really squares but oblongs. To
make it a true Greek cross we should require the additions that I have
indicated with dotted lines. Of course his solution produces a cross, but it is
not the symmetrical Greek variety required by the conditions of the puzzle.
My young friend thought his attempt was "near enough" to be correct; but if
he bought a penny apple with a sixpence he probably would not have
thought it "near enough" if he had been given only fourpence change. As
the reader advances he will realize the importance of this question of
exactitude.

In these cutting-out puzzles it is necessary not only to get the directions of
the cutting lines as correct as possible, but to remember that these lines
have no width. If after cutting up one of the crosses in a manner indicated in
these articles you find that the pieces do not exactly fit to form a square,
you may be certain that the fault is entirely your own. Either your cross was
not exactly drawn, or your cuts were not made quite in the right directions,
or (if you used wood and a fret-saw) your saw was not sufficiently fine. If
you cut out the puzzles in paper with scissors, or in cardboard with a
penknife, no material is lost; but with a saw, however fine, there is a certain
loss. In the case of most puzzles this slight loss is not sufficient to be
appreciable, if the puzzle is cut out on a large scale, but there have been
instances where I have found it desirable to draw and cut out each part
separately—not from one diagram—in order to produce a perfect result.

Now for another puzzle. If you have cut out the five pieces indicated in Fig.
14, you will find that these can be put together so as to form the curious

cross shown in Fig. 23. So if I asked you to cut Fig. 24 into five pieces to
form either a square or two equal Greek crosses you would know how to do
it. You would make the cuts as in Fig. 23, and place them together as in
Figs. 14 and 15. But I want something better than that, and it is this. Cut
Fig. 24 into only four pieces that will fit together and form a square.

The solution to the puzzle is shown in Figs. 25 and 26. The direction of the
cut dividing A and C in the first diagram is very obvious, and the second cut
is made at right angles to it. That the four pieces should fit together and
form a square will surprise the novice, who will do well to study the puzzle
with some care, as it is most instructive.
I will now explain the beautiful rule by which we determine the size of a
square that shall have the same area as a Greek cross, for it is applicable,
and necessary, to the solution of almost every dissection puzzle that we
meet with. It was first discovered by the philosopher Pythagoras, who died
500 B.C., and is the 47th proposition of Euclid. The young reader who
knows nothing of the elements of geometry will get some idea of the
fascinating character of that science. The triangle ABC in Fig. 27 is what
we call a right-angled triangle, because the side BC is at right angles to the
side AB. Now if we build up a square on each side of the triangle, the
squares on AB and BC will together be exactly equal to the square on the
long side AC, which we call the hypotenuse. This is proved in the case I
have given by subdividing the three squares into cells of equal dimensions.

It will be seen that 9 added to 16 equals 25, the number of cells in the large
square. If you make triangles with the sides 5, 12 and 13, or with 8, 15 and
17, you will get similar arithmetical proofs, for these are all "rational" right-
angled triangles, but the law is equally true for all cases. Supposing we cut
off the lower arm of a Greek cross and place it to the left of the upper arm,
as in Fig. 28, then the square on EF added to the square on DE exactly
equals a square on DF. Therefore we know that the square of DF will
contain the same area as the cross. This fact we have proved practically by
the solutions of the earlier puzzles of this series. But whatever length we
give to DE and EF, we can never give the exact length of DF in numbers,
because the triangle is not a "rational" one. But the law is none the less
geometrically true.

Now look at Fig. 29, and you will see an elegant method for cutting a piece
of wood of the shape of two squares (of any relative dimensions) into three
pieces that will fit together and form a single square. If you mark off the
distance ab equal to the side cd the directions of the cuts are very evident.
From what we have just been considering, you will at once see why bc must

be the length of the side of the new square. Make the experiment as often as
you like, taking different relative proportions for the two squares, and you
will find the rule always come true. If you make the two squares of exactly
the same size, you will see that the diagonal of any square is always the side
of a square that is twice the size. All this, which is so simple that anybody
can understand it, is very essential to the solving of cutting-out puzzles. It is
in fact the key to most of them. And it is all so beautiful that it seems a pity
that it should not be familiar to everybody.

We will now go one step further and deal with the half-square. Take a
square and cut it in half diagonally. Now try to discover how to cut this
triangle into four pieces that will form a Greek cross. The solution is shown
in Figs. 31 and 32. In this case it will be seen that we divide two of the sides
of the triangle into three equal parts and the long side into four equal parts.
Then the direction of the cuts will be easily found. It is a pretty puzzle, and
a little more difficult than some of the others that I have given. It should be
noted again that it would have been much easier to locate the cuts in the
reverse puzzle of cutting the cross to form a half-square triangle.

Another ideal that the puzzle maker always keeps in mind is to contrive that
there shall, if possible, be only one correct solution. Thus, in the case of the
first puzzle, if we only require that a Greek cross shall be cut into four
pieces to form a square, there is, as I have shown, an infinite number of
different solutions. It makes a better puzzle to add the condition that all the
four pieces shall be of the same size and shape, because it can then be
solved in only one way, as in Figs. 8 and 9. In this way, too, a puzzle that is
too easy to be interesting may be improved by such an addition. Let us take
an example. We have seen in Fig. 28 that Fig. 33 can be cut into two pieces
to form a Greek cross. I suppose an intelligent child would do it in five
minutes. But suppose we say that the puzzle has to be solved with a piece of
wood that has a bad knot in the position shown in Fig. 33—a knot that we
must not attempt to cut through—then a solution in two pieces is barred out,
and it becomes a more interesting puzzle to solve it in three pieces. I have
shown in Figs. 33 and 34 one way of doing this, and it will be found
entertaining to discover other ways of doing it. Of course I could bar out all
these other ways by introducing more knots, and so reduce the puzzle to a
single solution, but it would then be overloaded with conditions.
And this brings us to another point in seeking the ideal. Do not overload
your conditions, or you will make your puzzle too complex to be
interesting. The simpler the conditions of a puzzle are, the better. The
solution may be as complex and difficult as you like, or as happens, but the
conditions ought to be easily understood, or people will not attempt a
solution.

If the reader were now asked "to cut a half-square into as few pieces as
possible to form a Greek cross," he would probably produce our solution,
Figs. 31-32, and confidently claim that he had solved the puzzle correctly.
In this way he would be wrong, because it is not now stated that the square
is to be divided diagonally. Although we should always observe the exact
conditions of a puzzle we must not read into it conditions that are not there.
Many puzzles are based entirely on the tendency that people have to do this.
The very first essential in solving a puzzle is to be sure that you understand
the exact conditions. Now, if you divided your square in half so as to
produce Fig. 35 it is possible to cut it into as few as three pieces to form a
Greek cross. We thus save a piece.

I give another puzzle in Fig. 36. The dotted lines are added merely to show
the correct proportions of the figure—a square of 25 cells with the four
corner cells cut out. The puzzle is to cut this figure into five pieces that will
form a Greek cross (entire) and a square.
The solution to the first of the two puzzles last given—to cut a rectangle of
the shape of a half-square into three pieces that will form a Greek cross—is
shown in Figs. 37 and 38. It will be seen that we divide the long sides of the
oblong into six equal parts and the short sides into three equal parts, in
order to get the points that will indicate the direction of the cuts. The reader
should compare this solution with some of the previous illustrations. He
will see, for example, that if we continue the cut that divides B and C in the
cross, we get Fig. 15.

The other puzzle, like the one illustrated in Figs. 12 and 13, will show how
useful a little arithmetic may sometimes prove to be in the solution of
dissection puzzles. There are twenty-one of those little square cells into
which our figure is subdivided, from which we have to form both a square
and a Greek cross. Now, as the cross is built up of five squares, and 5 from
21 leaves 16—a square number—we ought easily to be led to the solution
shown in Fig. 39. It will be seen that the cross is cut out entire, while the
four remaining pieces form the square in Fig. 40.

Of course a half-square rectangle is the same as a double square, or two
equal squares joined together. Therefore, if you want to solve the puzzle of
cutting a Greek cross into four pieces to form two separate squares of the
same size, all you have to do is to continue the short cut in Fig. 38 right
across the cross, and you will have four pieces of the same size and shape.

Now divide Fig. 37 into two equal squares by a horizontal cut midway and
you will see the four pieces forming the two squares.

Cut a Greek cross into five pieces that will form two separate squares, one
of which shall contain half the area of one of the arms of the cross. In
further illustration of what I have already written, if the two squares of the
same size A B C D and B C F E, in Fig. 41, are cut in the manner indicated
by the dotted lines, the four pieces will form the large square A G E C. We
thus see that the diagonal A C is the side of a square twice the size of A B C
D. It is also clear that half the diagonal of any square is equal to the side of
a square of half the area. Therefore, if the large square in the diagram is one
of the arms of your cross, the small square is the size of one of the squares
required in the puzzle.

The solution is shown in Figs. 42 and 43. It will be seen that the small
square is cut out whole and the large square composed of the four pieces B,
C, D, and E. After what I have written, the reader will have no difficulty in
seeing that the square A is half the size of one of the arms of the cross,
because the length of the diagonal of the former is clearly the same as the
side of the latter. The thing is now self-evident. I have thus tried to show
that some of these puzzles that many people are apt to regard as quite
wonderful and bewildering, are really not difficult if only we use a little
thought and judgment. In conclusion of this particular subject I will give
four Greek cross puzzles, with detached solutions.

142.—THE SILK PATCHWORK.
The lady members of the Wilkinson family had made a simple patchwork
quilt, as a small Christmas present, all composed of square pieces of the
same size, as shown in the illustration. It only lacked the four corner pieces
to make it complete. Somebody pointed out to them that if you unpicked the
Greek cross in the middle and then cut the stitches along the dark joins, the
four pieces all of the same size and shape would fit together and form a
square. This the reader knows, from the solution in Fig. 39, is quite easily
done. But George Wilkinson suddenly suggested to them this poser. He
said, "Instead of picking out the cross entire, and forming the square from
four equal pieces, can you cut out a square entire and four equal pieces that
will form a perfect Greek cross?" The puzzle is, of course, now quite easy.

143.—TWO CROSSES FROM ONE.
Cut a Greek cross into five pieces that will form two such crosses, both of
the same size. The solution of this puzzle is very beautiful.

144.—THE CROSS AND THE TRIANGLE.
Cut a Greek cross into six pieces that will form an equilateral triangle. This
is another hard problem, and I will state here that a solution is practically
impossible without a previous knowledge of my method of transforming an
equilateral triangle into a square (see No. 26, "Canterbury Puzzles").

145.—THE FOLDED CROSS.
Cut out of paper a Greek cross; then so fold it that with a single straight cut
of the scissors the four pieces produced will form a square.

VARIOUS DISSECTION PUZZLES.
We will now consider a small miscellaneous selection of cutting-out
puzzles, varying in degrees of difficulty.

146.—AN EASY DISSECTION PUZZLE.

First, cut out a piece of paper or cardboard of the shape shown in the
illustration. It will be seen at once that the proportions are simply those of a
square attached to half of another similar square, divided diagonally. The
puzzle is to cut it into four pieces all of precisely the same size and shape.

147.—AN EASY SQUARE PUZZLE.

If you take a rectangular piece of cardboard, twice as long as it is broad, and
cut it in half diagonally, you will get two of the pieces shown in the
illustration. The puzzle is with five such pieces of equal size to form a
square. One of the pieces may be cut in two, but the others must be used
intact.

148.—THE BUN PUZZLE.

The three circles represent three buns, and it is simply required to show
how these may be equally divided among four boys. The buns must be
regarded as of equal thickness throughout and of equal thickness to each
other. Of course, they must be cut into as few pieces as possible. To
simplify it I will state the rather surprising fact that only five pieces are

necessary, from which it will be seen that one boy gets his share in two
pieces and the other three receive theirs in a single piece. I am aware that
this statement "gives away" the puzzle, but it should not destroy its interest
to those who like to discover the "reason why."

149.—THE CHOCOLATE SQUARES.

Here is a slab of chocolate, indented at the dotted lines so that the twenty
squares can be easily separated. Make a copy of the slab in paper or
cardboard and then try to cut it into nine pieces so that they will form four
perfect squares all of exactly the same size.

150.—DISSECTING A MITRE.

The figure that is perplexing the carpenter in the illustration represents a
mitre. It will be seen that its proportions are those of a square with one
quarter removed. The puzzle is to cut it into five pieces that will fit together
and form a perfect square. I show an attempt, published in America, to
perform the feat in four pieces, based on what is known as the "step
principle," but it is a fallacy.

We are told first to cut oft the pieces 1 and 2 and pack them into the
triangular space marked off by the dotted line, and so form a rectangle.
So far, so good. Now, we are directed to apply the old step principle, as
shown, and, by moving down the piece 4 one step, form the required
square. But, unfortunately, it does not produce a square: only an oblong.
Call the three long sides of the mitre 84 in. each. Then, before cutting the
steps, our rectangle in three pieces will be 84×63. The steps must be 10½
in. in height and 12 in. in breadth. Therefore, by moving down a step we
reduce by 12 in. the side 84 in. and increase by 10½ in. the side 63 in.
Hence our final rectangle must be 72 in. × 73½ in., which certainly is not a
square! The fact is, the step principle can only be applied to rectangles with
sides of particular relative lengths. For example, if the shorter side in this
case were 615/7 (instead of 63), then the step method would apply. For the
steps would then be 102/7 in. in height and 12 in. in breadth. Note that
615/7 × 84= the square of 72. At present no solution has been found in four
pieces, and I do not believe one possible.

151.—THE JOINER'S PROBLEM.
I have often had occasion to remark on the practical utility of puzzles,
arising out of an application to the ordinary affairs of life of the little tricks
and "wrinkles" that we learn while solving recreation problems.

The joiner, in the illustration, wants to cut the piece of wood into as few
pieces as possible to form a square table-top, without any waste of material.
How should he go to work? How many pieces would you require?

152.—ANOTHER JOINER'S PROBLEM.

A joiner had two pieces of wood of the shapes and relative proportions
shown in the diagram. He wished to cut them into as few pieces as possible
so that they could be fitted together, without waste, to form a perfectly
square table-top. How should he have done it? There is no necessity to give
measurements, for if the smaller piece (which is half a square) be made a
little too large or a little too small it will not affect the method of solution.

153.—A CUTTING-OUT PUZZLE.
Here is a little cutting-out poser. I take a strip of paper, measuring five
inches by one inch, and, by cutting it into five pieces, the parts fit together
and form a square, as shown in the illustration. Now, it is quite an
interesting puzzle to discover how we can do this in only four pieces.

154.—MRS. HOBSON'S HEARTHRUG.

Mrs. Hobson's boy had an accident when playing with the fire, and burnt
two of the corners of a pretty hearthrug. The damaged corners have been cut
away, and it now has the appearance and proportions shown in my diagram.
How is Mrs. Hobson to cut the rug into the fewest possible pieces that will
fit together and form a perfectly square rug? It will be seen that the rug is in
the proportions 36 × 27 (it does not matter whether we say inches or yards),
and each piece cut away measured 12 and 6 on the outside.

155.—THE PENTAGON AND SQUARE.
I wonder how many of my readers, amongst those who have not given any
close attention to the elements of geometry, could draw a regular pentagon,
or five-sided figure, if they suddenly required to do so. A regular hexagon,
or six-sided figure, is easy enough, for everybody knows that all you have
to do is to describe a circle and then, taking the radius as the length of one
of the sides, mark off the six points round the circumference. But a
pentagon is quite another matter. So, as my puzzle has to do with the cutting
up of a regular pentagon, it will perhaps be well if I first show my less

experienced readers how this figure is to be correctly drawn. Describe a
circle and draw the two lines H B and D G, in the diagram, through the
centre at right angles. Now find the point A, midway between C and B.
Next place the point of your compasses at A and with the distance A D
describe the arc cutting H B at E. Then place the point of your compasses at
D and with the distance D E describe the arc cutting the circumference at F.
Now, D F is one of the sides of your pentagon, and you have simply to
mark off the other sides round the circle. Quite simple when you know how,
but otherwise somewhat of a poser.

Having formed your pentagon, the puzzle is to cut it into the fewest
possible pieces that will fit together and form a perfect square.

156.—THE DISSECTED TRIANGLE.

A good puzzle is that which the gentleman in the illustration is showing to
his friends. He has simply cut out of paper an equilateral triangle—that is, a
triangle with all its three sides of the same length. He proposes that it shall
be cut into five pieces in such a way that they will fit together and form
either two or three smaller equilateral triangles, using all the material in
each case. Can you discover how the cuts should be made?
Remember that when you have made your five pieces, you must be able, as
desired, to put them together to form either the single original triangle or to
form two triangles or to form three triangles—all equilateral.

157.—THE TABLE-TOP AND STOOLS.
I have frequently had occasion to show that the published answers to a great
many of the oldest and most widely known puzzles are either quite
incorrect or capable of improvement. I propose to consider the old poser of
the table-top and stools that most of my readers have probably seen in some
form or another in books compiled for the recreation of childhood.

The story is told that an economical and ingenious schoolmaster once
wished to convert a circular table-top, for which he had no use, into seats
for two oval stools, each with a hand-hole in the centre. He instructed the
carpenter to make the cuts as in the illustration and then join the eight
pieces together in the manner shown. So impressed was he with the
ingenuity of his performance that he set the puzzle to his geometry class as
a little study in dissection. But the remainder of the story has never been
published, because, so it is said, it was a characteristic of the principals of
academies that they would never admit that they could err. I get my
information from a descendant of the original boy who had most reason to
be interested in the matter.
The clever youth suggested modestly to the master that the hand-holes were
too big, and that a small boy might perhaps fall through them. He therefore
proposed another way of making the cuts that would get over this objection.
For his impertinence he received such severe chastisement that he became
convinced that the larger the hand-hole in the stools the more comfortable
might they be.

Now what was the method the boy proposed?
Can you show how the circular table-top may be cut into eight pieces that
will fit together and form two oval seats for stools (each of exactly the same
size and shape) and each having similar hand-holes of smaller dimensions
than in the case shown above? Of course, all the wood must be used.

158.—THE GREAT MONAD.

Here is a symbol of tremendous antiquity which is worthy of notice. It is
borne on the Korean ensign and merchant flag, and has been adopted as a
trade sign by the Northern Pacific Railroad Company, though probably few
are aware that it is the Great Monad, as shown in the sketch below. This
sign is to the Chinaman what the cross is to the Christian. It is the sign of
Deity and eternity, while the two parts into which the circle is divided are
called the Yin and the Yan—the male and female forces of nature. A writer
on the subject more than three thousand years ago is reported to have said
in reference to it: "The illimitable produces the great extreme. The great
extreme produces the two principles. The two principles produce the four
quarters, and from the four quarters we develop the quadrature of the eight
diagrams of Feuh-hi." I hope readers will not ask me to explain this, for I
have not the slightest idea what it means. Yet I am persuaded that for ages
the symbol has had occult and probably mathematical meanings for the
esoteric student.

I will introduce the Monad in its elementary form. Here are three easy
questions respecting this great symbol:—
(I.) Which has the greater area, the inner circle containing the Yin and the
Yan, or the outer ring?
(II.) Divide the Yin and the Yan into four pieces of the same size and shape
by one cut.
(III.) Divide the Yin and the Yan into four pieces of the same size, but
different shape, by one straight cut.

159.—THE SQUARE OF VENEER.
The following represents a piece of wood in my possession, 5 in. square. By
markings on the surface it is divided into twenty-five square inches. I want
to discover a way of cutting this piece of wood into the fewest possible
pieces that will fit together and form two perfect squares of different sizes
and of known dimensions. But, unfortunately, at every one of the sixteen
intersections of the cross lines a small nail has been driven in at some time
or other, and my fret-saw will be injured if it comes in contact with any of
these. I have therefore to find a method of doing the work that will not
necessitate my cutting through any of those sixteen points. How is it to be
done? Remember, the exact dimensions of the two squares must be given.

160.—THE TWO HORSESHOES.
Why horseshoes should be considered "lucky" is one of those things which
no man can understand. It is a very old superstition, and John Aubrey
(1626-1700) says, "Most houses at the West End of London have a
horseshoe on the threshold." In Monmouth Street there were seventeen in
1813 and seven so late as 1855. Even Lord Nelson had one nailed to the
mast of the ship Victory. To-day we find it more conducive to "good luck"
to see that they are securely nailed on the feet of the horse we are about to
drive.
Nevertheless, so far as the horseshoe, like the Swastika and other emblems
that I have had occasion at times to deal with, has served to symbolize
health, prosperity, and goodwill towards men, we may well treat it with a
certain amount of respectful interest. May there not, moreover, be some
esoteric or lost mathematical mystery concealed in the form of a horseshoe?
I have been looking into this matter, and I wish to draw my readers'

attention to the very remarkable fact that the pair of horseshoes shown in
my illustration are related in a striking and beautiful manner to the circle,
which is the symbol of eternity. I present this fact in the form of a simple
problem, so that it may be seen how subtly this relation has been concealed
for ages and ages. My readers will, I know, be pleased when they find the
key to the mystery.

Cut out the two horseshoes carefully round the outline and then cut them
into four pieces, all different in shape, that will fit together and form a
perfect circle. Each shoe must be cut into two pieces and all the part of the
horse's hoof contained within the outline is to be used and regarded as part
of the area.

161.—THE BETSY ROSS PUZZLE.
A correspondent asked me to supply him with the solution to an old puzzle
that is attributed to a certain Betsy Ross, of Philadelphia, who showed it to
George Washington. It consists in so folding a piece of paper that with one
clip of the scissors a five-pointed star of Freedom may be produced.
Whether the story of the puzzle's origin is a true one or not I cannot say, but
I have a print of the old house in Philadelphia where the lady is said to have
lived, and I believe it still stands there. But my readers will doubtless be
interested in the little poser.

Take a circular piece of paper and so fold it that with one cut of the scissors
you can produce a perfect five-pointed star.

162.—THE CARDBOARD CHAIN.

Can you cut this chain out of a piece of cardboard without any join
whatever? Every link is solid; without its having been split and afterwards
joined at any place. It is an interesting old puzzle that I learnt as a child, but
I have no knowledge as to its inventor.

163.—THE PAPER BOX.
It may be interesting to introduce here, though it is not strictly a puzzle, an
ingenious method for making a paper box.
Take a square of stout paper and by successive foldings make all the creases
indicated by the dotted lines in the illustration. Then cut away the eight little
triangular pieces that are shaded, and cut through the paper along the dark
lines. The second illustration shows the box half folded up, and the reader
will have no difficulty in effecting its completion. Before folding up, the
reader might cut out the circular piece indicated in the diagram, for a
purpose I will now explain.

This box will be found to serve excellently for the production of vortex
rings. These rings, which were discussed by Von Helmholtz in 1858, are
most interesting, and the box (with the hole cut out) will produce them to
perfection. Fill the box with tobacco smoke by blowing it gently through
the hole. Now, if you hold it horizontally, and softly tap the side that is
opposite to the hole, an immense number of perfect rings can be produced
from one mouthful of smoke. It is best that there should be no currents of
air in the room. People often do not realise that these rings are formed in the
air when no smoke is used. The smoke only makes them visible. Now, one
of these rings, if properly directed on its course, will travel across the room
and put out the flame of a candle, and this feat is much more striking if you
can manage to do it without the smoke. Of course, with a little practice, the
rings may be blown from the mouth, but the box produces them in much
greater perfection, and no skill whatever is required. Lord Kelvin
propounded the theory that matter may consist of vortex rings in a fluid that
fills all space, and by a development of the hypothesis he was able to
explain chemical combination.

164.—THE POTATO PUZZLE.
Take a circular slice of potato, place it on the table, and see into how large a
number of pieces you can divide it with six cuts of a knife. Of course you
must not readjust the pieces or pile them after a cut. What is the greatest
number of pieces you can make?

The illustration shows how to make sixteen pieces. This can, of course, be
easily beaten.

165.—THE SEVEN PIGS.

Here is a little puzzle that was put to one of the sons of Erin the other day
and perplexed him unduly, for it is really quite easy. It will be seen from the
illustration that he was shown a sketch of a square pen containing seven
pigs. He was asked how he would intersect the pen with three straight
fences so as to enclose every pig in a separate sty. In other words, all you
have to do is to take your pencil and, with three straight strokes across the
square, enclose each pig separately. Nothing could be simpler.
The Irishman complained that the pigs would not keep still while he was
putting up the fences. He said that they would all flock together, or one
obstinate beast would go into a corner and flock all by himself. It was
pointed out to him that for the purposes of the puzzle the pigs were
stationary. He answered that Irish pigs are not stationery—they are pork.
Being persuaded to make the attempt, he drew three lines, one of which cut
through a pig. When it was explained that this is not allowed, he protested

that a pig was no use until you cut its throat. "Begorra, if it's bacon ye want
without cutting your pig, it will be all gammon." We will not do the
Irishman the injustice of suggesting that the miserable pun was intentional.
However, he failed to solve the puzzle. Can you do it?

166.—THE LANDOWNER'S FENCES.

The landowner in the illustration is consulting with his bailiff over a rather
puzzling little question. He has a large plan of one of his fields, in which
there are eleven trees. Now, he wants to divide the field into just eleven
enclosures by means of straight fences, so that every enclosure shall contain
one tree as a shelter for his cattle. How is he to do it with as few fences as

possible? Take your pencil and draw straight lines across the field until you
have marked off the eleven enclosures (and no more), and then see how
many fences you require. Of course the fences may cross one another.

167.—THE WIZARD'S CATS.

A wizard placed ten cats inside a magic circle as shown in our illustration,
and hypnotized them so that they should remain stationary during his
pleasure. He then proposed to draw three circles inside the large one, so that
no cat could approach another cat without crossing a magic circle. Try to
draw the three circles so that every cat has its own enclosure and cannot
reach another cat without crossing a line.

168.—THE CHRISTMAS PUDDING.

"Speaking of Christmas puddings," said the host, as he glanced at the
imposing delicacy at the other end of the table. "I am reminded of the fact
that a friend gave me a new puzzle the other day respecting one. Here it is,"
he added, diving into his breast pocket.
"'Problem: To find the contents,' I suppose," said the Eton boy.
"No; the proof of that is in the eating. I will read you the conditions."

"'Cut the pudding into two parts, each of exactly the same size and shape,
without touching any of the plums. The pudding is to be regarded as a flat
disc, not as a sphere.'"
"Why should you regard a Christmas pudding as a disc? And why should
any reasonable person ever wish to make such an accurate division?" asked
the cynic.
"It is just a puzzle—a problem in dissection." All in turn had a look at the
puzzle, but nobody succeeded in solving it. It is a little difficult unless you

are acquainted with the principle involved in the making of such puddings,
but easy enough when you know how it is done.

169.—A TANGRAM PARADOX.
Many pastimes of great antiquity, such as chess, have so developed and
changed down the centuries that their original inventors would scarcely
recognize them. This is not the case with Tangrams, a recreation that
appears to be at least four thousand years old, that has apparently never
been dormant, and that has not been altered or "improved upon" since the
legendary Chinaman Tan first cut out the seven pieces shown in Diagram I.
If you mark the point B, midway between A and C, on one side of a square
of any size, and D, midway between C and E, on an adjoining side, the
direction of the cuts is too obvious to need further explanation. Every
design in this article is built up from the seven pieces of blackened
cardboard. It will at once be understood that the possible combinations are
infinite.

The late Mr. Sam Loyd, of New York, who published a small book of very
ingenious designs, possessed the manuscripts of the late Mr. Challenor, who
made a long and close study of Tangrams. This gentleman, it is said, records
that there were originally seven books of Tangrams, compiled in China two
thousand years before the Christian era. These books are so rare that, after

forty years' residence in the country, he only succeeded in seeing perfect
copies of the first and seventh volumes with fragments of the second.
Portions of one of the books, printed in gold leaf upon parchment, were
found in Peking by an English soldier and sold for three hundred pounds.
A few years ago a little book came into my possession, from the library of
the late Lewis Carroll, entitled The Fashionable Chinese Puzzle. It contains
three hundred and twenty-three Tangram designs, mostly nondescript
geometrical figures, to be constructed from the seven pieces. It was
"Published by J. and E. Wallis, 42 Skinner Street, and J. Wallis, Jun.,
Marine Library, Sidmouth" (South Devon). There is no date, but the
following note fixes the time of publication pretty closely: "This ingenious
contrivance has for some time past been the favourite amusement of the ex-
Emperor Napoleon, who, being now in a debilitated state and living very
retired, passes many hours a day in thus exercising his patience and
ingenuity." The reader will find, as did the great exile, that much
amusement, not wholly uninstructive, may be derived from forming the
designs of others. He will find many of the illustrations to this article quite
easy to build up, and some rather difficult. Every picture may thus be
regarded as a puzzle.
But it is another pastime altogether to create new and original designs of a
pictorial character, and it is surprising what extraordinary scope the
Tangrams afford for producing pictures of real life—angular and often
grotesque, it is true, but full of character. I give an example of a recumbent
figure (2) that is particularly graceful, and only needs some slight reduction
of its angularities to produce an entirely satisfactory outline.

As I have referred to the author of Alice in Wonderland, I give also my
designs of the March Hare (3) and the Hatter (4). I also give an attempt at
Napoleon (5), and a very excellent Red Indian with his Squaw by Mr. Loyd
(6 and 7). A large number of other designs will be found in an article by me
in The Strand Magazine for November, 1908.

On the appearance of this magazine article, the late Sir James Murray, the
eminent philologist, tried, with that amazing industry that characterized all
his work, to trace the word "tangram" to its source. At length he wrote as
follows:—"One of my sons is a professor in the Anglo-Chinese college at
Tientsin. Through him, his colleagues, and his students, I was able to make
inquiries as to the alleged Tan among Chinese scholars. Our Chinese
professor here (Oxford) also took an interest in the matter and obtained
information from the secretary of the Chinese Legation in London, who is a
very eminent representative of the Chinese literati."
"The result has been to show that the man Tan, the god Tan, and the 'Book
of Tan' are entirely unknown to Chinese literature, history, or tradition. By
most of the learned men the name, or allegation of the existence, of these
had never been heard of. The puzzle is, of course, well known. It is called in
Chinese ch'i ch'iao t'u; literally, 'seven-ingenious-plan' or 'ingenious-puzzle
figure of seven pieces.' No name approaching 'tangram,' or even 'tan,'
occurs in Chinese, and the only suggestions for the latter were the Chinese
t'an, 'to extend'; or t'ang, Cantonese dialect for 'Chinese.' It was suggested
that probably some American or Englishman who knew a little Chinese or
Cantonese, wanting a name for the puzzle, might concoct one out of one of

these words and the European ending 'gram.' I should say the name
'tangram' was probably invented by an American some little time before
1864 and after 1847, but I cannot find it in print before the 1864 edition of
Webster. I have therefore had to deal very shortly with the word in the
dictionary, telling what it is applied to and what conjectures or guesses have
been made at the name, and giving a few quotations, one from your own
article, which has enabled me to make more of the subject than I could
otherwise have done."
Several correspondents have informed me that they possess, or had
possessed, specimens of the old Chinese books. An American gentleman
writes to me as follows:—"I have in my possession a book made of tissue
paper, printed in black (with a Chinese inscription on the front page),
containing over three hundred designs, which belongs to the box of
'tangrams,' which I also own. The blocks are seven in number, made of
mother-of-pearl, highly polished and finely engraved on either side. These
are contained in a rosewood box 21/8 in. square. My great uncle, ——, was
one of the first missionaries to visit China. This box and book, along with
quite a collection of other relics, were sent to my grandfather and descended
to myself."

My correspondent kindly supplied me with rubbings of the Tangrams, from
which it is clear that they are cut in the exact proportions that I have

indicated. I reproduce the Chinese inscription (8) for this reason. The owner
of the book informs me that he has submitted it to a number of Chinamen in
the United States and offered as much as a dollar for a translation. But they
all steadfastly refused to read the words, offering the lame excuse that the
inscription is Japanese. Natives of Japan, however, insist that it is Chinese.
Is there something occult and esoteric about Tangrams, that it is so difficult
to lift the veil? Perhaps this page will come under the eye of some reader
acquainted with the Chinese language, who will supply the required
translation, which may, or may not, throw a little light on this curious
question.

By using several sets of Tangrams at the same time we may construct more
ambitious pictures. I was advised by a friend not to send my picture, "A
Game of Billiards" (9), to the Academy. He assured me that it would not be
accepted because the "judges are so hide-bound by convention." Perhaps he
was right, and it will be more appreciated by Post-impressionists and
Cubists. The players are considering a very delicate stroke at the top of the
table. Of course, the two men, the table, and the clock are formed from four
sets of Tangrams. My second picture is named "The Orchestra" (10), and it
was designed for the decoration of a large hall of music. Here we have the
conductor, the pianist, the fat little cornet-player, the left-handed player of
the double-bass, whose attitude is life-like, though he does stand at an
unusual distance from his instrument, and the drummer-boy, with his
imposing music-stand. The dog at the back of the pianoforte is not howling:
he is an appreciative listener.

One remarkable thing about these Tangram pictures is that they suggest to
the imagination such a lot that is not really there. Who, for example, can
look for a few minutes at Lady Belinda (11) and the Dutch girl (12) without
soon feeling the haughty expression in the one case and the arch look in the
other? Then look again at the stork (13), and see how it is suggested to the
mind that the leg is actually much more slender than any one of the pieces
employed. It is really an optical illusion. Again, notice in the case of the
yacht (14) how, by leaving that little angular point at the top, a complete
mast is suggested. If you place your Tangrams together on white paper so
that they do not quite touch one another, in some cases the effect is
improved by the white lines; in other cases it is almost destroyed.

Finally, I give an example from the many curious paradoxes that one
happens upon in manipulating Tangrams. I show designs of two dignified
individuals (15 and 16) who appear to be exactly alike, except for the fact
that one has a foot and the other has not. Now, both of these figures are
made from the same seven Tangrams. Where does the second man get his
foot from?

PATCHWORK PUZZLES.
"Of shreds and patches."—Hamlet, iii. 4.

170.—THE CUSHION COVERS.

The above represents a square of brocade. A lady wishes to cut it in four
pieces so that two pieces will form one perfectly square cushion top, and
the remaining two pieces another square cushion top. How is she to do it?
Of course, she can only cut along the lines that divide the twenty-five
squares, and the pattern must "match" properly without any irregularity
whatever in the design of the material. There is only one way of doing it.
Can you find it?

171.—THE BANNER PUZZLE.

A Lady had a square piece of bunting with two lions on it, of which the
illustration is an exactly reproduced reduction. She wished to cut the stuff
into pieces that would fit together and form two square banners with a lion
on each banner. She discovered that this could be done in as few as four
pieces. How did she manage it? Of course, to cut the British Lion would be
an unpardonable offence, so you must be careful that no cut passes through
any portion of either of them. Ladies are informed that no allowance
whatever has to be made for "turnings," and no part of the material may be
wasted. It is quite a simple little dissection puzzle if rightly attacked.
Remember that the banners have to be perfect squares, though they need not
be both of the same size.

172.—MRS. SMILEY'S CHRISTMAS PRESENT.

Mrs. Smiley's expression of pleasure was sincere when her six
granddaughters sent to her, as a Christmas present, a very pretty patchwork
quilt, which they had made with their own hands. It was constructed of
square pieces of silk material, all of one size, and as they made a large quilt
with fourteen of these little squares on each side, it is obvious that just 196
pieces had been stitched into it. Now, the six granddaughters each
contributed a part of the work in the form of a perfect square (all six
portions being different in size), but in order to join them up to form the
square quilt it was necessary that the work of one girl should be unpicked
into three separate pieces. Can you show how the joins might have been
made? Of course, no portion can be turned over.

173.—MRS. PERKINS'S QUILT.

It will be seen that in this case the square patchwork quilt is built up of 169
pieces. The puzzle is to find the smallest possible number of square portions
of which the quilt could be composed and show how they might be joined
together. Or, to put it the reverse way, divide the quilt into as few square
portions as possible by merely cutting the stitches.

174.—THE SQUARES OF BROCADE.

I happened to be paying a call at the house of a lady, when I took up from a
table two lovely squares of brocade. They were beautiful specimens of
Eastern workmanship—both of the same design, a delicate chequered
pattern.
"Are they not exquisite?" said my friend. "They were brought to me by a
cousin who has just returned from India. Now, I want you to give me a little
assistance. You see, I have decided to join them together so as to make one
large square cushion-cover. How should I do this so as to mutilate the
material as little as possible? Of course I propose to make my cuts only
along the lines that divide the little chequers."

I cut the two squares in the manner desired into four pieces that would fit
together and form another larger square, taking care that the pattern should
match properly, and when I had finished I noticed that two of the pieces
were of exactly the same area; that is, each of the two contained the same
number of chequers. Can you show how the cuts were made in accordance
with these conditions?

175.—ANOTHER PATCHWORK PUZZLE.

A lady was presented, by two of her girl friends, with the pretty pieces of
silk patchwork shown in our illustration. It will be seen that both pieces are
made up of squares all of the same size—one 12x12 and the other 5x5. She
proposes to join them together and make one square patchwork quilt,
13x13, but, of course, she will not cut any of the material—merely cut the
stitches where necessary and join together again. What perplexes her is this.
A friend assures her that there need be no more than four pieces in all to
join up for the new quilt. Could you show her how this little needlework
puzzle is to be solved in so few pieces?

176.—LINOLEUM CUTTING.

The diagram herewith represents two separate pieces of linoleum. The
chequered pattern is not repeated at the back, so that the pieces cannot be
turned over. The puzzle is to cut the two squares into four pieces so that
they shall fit together and form one perfect square 10×10, so that the pattern
shall properly match, and so that the larger piece shall have as small a
portion as possible cut from it.

177.—ANOTHER LINOLEUM PUZZLE.

Can you cut this piece of linoleum into four pieces that will fit together and
form a perfect square? Of course the cuts may only be made along the lines.

VARIOUS GEOMETRICAL PUZZLES.
"So various are the tastes of men."
MARK AKENSIDE.

178.—THE CARDBOARD BOX.
This puzzle is not difficult, but it will be found entertaining to discover the
simple rule for its solution. I have a rectangular cardboard box. The top has
an area of 120 square inches, the side 96 square inches, and the end 80
square inches. What are the exact dimensions of the box?

179.—STEALING THE BELL-ROPES.
Two men broke into a church tower one night to steal the bell-ropes. The
two ropes passed through holes in the wooden ceiling high above them, and
they lost no time in climbing to the top. Then one man drew his knife and
cut the rope above his head, in consequence of which he fell to the floor and
was badly injured. His fellow-thief called out that it served him right for
being such a fool. He said that he should have done as he was doing, upon
which he cut the rope below the place at which he held on. Then, to his
dismay, he found that he was in no better plight, for, after hanging on as
long as his strength lasted, he was compelled to let go and fall beside his
comrade. Here they were both found the next morning with their limbs
broken. How far did they fall? One of the ropes when they found it was just
touching the floor, and when you pulled the end to the wall, keeping the
rope taut, it touched a point just three inches above the floor, and the wall
was four feet from the rope when it hung at rest. How long was the rope
from floor to ceiling?

180.—THE FOUR SONS.
Readers will recognize the diagram as a familiar friend of their youth. A
man possessed a square-shaped estate. He bequeathed to his widow the
quarter of it that is shaded off. The remainder was to be divided equitably
amongst his four sons, so that each should receive land of exactly the same
area and exactly similar in shape. We are shown how this was done. But the
remainder of the story is not so generally known. In the centre of the estate
was a well, indicated by the dark spot, and Benjamin, Charles, and David
complained that the division was not "equitable," since Alfred had access to
this well, while they could not reach it without trespassing on somebody
else's land. The puzzle is to show how the estate is to be apportioned so that
each son shall have land of the same shape and area, and each have access
to the well without going off his own land.

181.—THE THREE RAILWAY STATIONS.
As I sat in a railway carriage I noticed at the other end of the compartment a
worthy squire, whom I knew by sight, engaged in conversation with another
passenger, who was evidently a friend of his.
"How far have you to drive to your place from the railway station?" asked
the stranger.
"Well," replied the squire, "if I get out at Appleford, it is just the same
distance as if I go to Bridgefield, another fifteen miles farther on; and if I
changed at Appleford and went thirteen miles from there to Carterton, it
would still be the same distance. You see, I am equidistant from the three
stations, so I get a good choice of trains."
Now I happened to know that Bridgefield is just fourteen miles from
Carterton, so I amused myself in working out the exact distance that the
squire had to drive home whichever station he got out at. What was the
distance?

182.—THE GARDEN PUZZLE.
Professor Rackbrain tells me that he was recently smoking a friendly pipe
under a tree in the garden of a country acquaintance. The garden was
enclosed by four straight walls, and his friend informed him that he had
measured these and found the lengths to be 80, 45, 100, and 63 yards
respectively. "Then," said the professor, "we can calculate the exact area of
the garden." "Impossible," his host replied, "because you can get an infinite
number of different shapes with those four sides." "But you forget,"
Rackbrane said, with a twinkle in his eye, "that you told me once you had
planted this tree equidistant from all the four corners of the garden." Can
you work out the garden's area?

183.—DRAWING A SPIRAL.

If you hold the page horizontally and give it a quick rotary motion while
looking at the centre of the spiral, it will appear to revolve. Perhaps a good
many readers are acquainted with this little optical illusion. But the puzzle
is to show how I was able to draw this spiral with so much exactitude
without using anything but a pair of compasses and the sheet of paper on
which the diagram was made. How would you proceed in such
circumstances?

184.—HOW TO DRAW AN OVAL.
Can you draw a perfect oval on a sheet of paper with one sweep of the
compasses? It is one of the easiest things in the world when you know how.

185.—ST. GEORGE'S BANNER.

At a celebration of the national festival of St. George's Day I was
contemplating the familiar banner of the patron saint of our country. We all
know the red cross on a white ground, shown in our illustration. This is the
banner of St. George. The banner of St. Andrew (Scotland) is a white "St.
Andrew's Cross" on a blue ground. That of St. Patrick (Ireland) is a similar
cross in red on a white ground. These three are united in one to form our
Union Jack.
Now on looking at St. George's banner it occurred to me that the following
question would make a simple but pretty little puzzle. Supposing the flag
measures four feet by three feet, how wide must the arm of the cross be if it
is required that there shall be used just the same quantity of red and of white
bunting?

186.—THE CLOTHES LINE PUZZLE.

A boy tied a clothes line from the top of each of two poles to the base of the
other. He then proposed to his father the following question. As one pole
was exactly seven feet above the ground and the other exactly five feet,
what was the height from the ground where the two cords crossed one
another?

187.—THE MILKMAID PUZZLE.
Here is a little pastoral puzzle that the reader may, at first sight, be led into
supposing is very profound, involving deep calculations. He may even say
that it is quite impossible to give any answer unless we are told something
definite as to the distances. And yet it is really quite "childlike and bland."

In the corner of a field is seen a milkmaid milking a cow, and on the other
side of the field is the dairy where the extract has to be deposited. But it has
been noticed that the young woman always goes down to the river with her
pail before returning to the dairy. Here the suspicious reader will perhaps
ask why she pays these visits to the river. I can only reply that it is no
business of ours. The alleged milk is entirely for local consumption.

"Where are you going to, my pretty maid?"
"Down to the river, sir," she said.
"I'll not choose your dairy, my pretty maid."
"Nobody axed you, sir," she said.
If one had any curiosity in the matter, such an independent spirit would
entirely disarm one. So we will pass from the point of commercial morality
to the subject of the puzzle.
Draw a line from the milking-stool down to the river and thence to the door
of the dairy, which shall indicate the shortest possible route for the
milkmaid. That is all. It is quite easy to indicate the exact spot on the bank
of the river to which she should direct her steps if she wants as short a walk
as possible. Can you find that spot?

188.—THE BALL PROBLEM.

A stonemason was engaged the other day in cutting out a round ball for the
purpose of some architectural decoration, when a smart schoolboy came
upon the scene.
"Look here," said the mason, "you seem to be a sharp youngster, can you
tell me this? If I placed this ball on the level ground, how many other balls
of the same size could I lay around it (also on the ground) so that every ball
should touch this one?"
The boy at once gave the correct answer, and then put this little question to
the mason:—
"If the surface of that ball contained just as many square feet as its volume
contained cubic feet, what would be the length of its diameter?"
The stonemason could not give an answer. Could you have replied correctly
to the mason's and the boy's questions?

189.—THE YORKSHIRE ESTATES.

I was on a visit to one of the large towns of Yorkshire. While walking to the
railway station on the day of my departure a man thrust a hand-bill upon
me, and I took this into the railway carriage and read it at my leisure. It
informed me that three Yorkshire neighbouring estates were to be offered
for sale. Each estate was square in shape, and they joined one another at
their corners, just as shown in the diagram. Estate A contains exactly 370
acres, B contains 116 acres, and C 74 acres.
Now, the little triangular bit of land enclosed by the three square estates was
not offered for sale, and, for no reason in particular, I became curious as to
the area of that piece. How many acres did it contain?

190.—FARMER WURZEL'S ESTATE.

I will now present another land problem. The demonstration of the answer
that I shall give will, I think, be found both interesting and easy of
comprehension.
Farmer Wurzel owned the three square fields shown in the annexed plan,
containing respectively 18, 20, and 26 acres. In order to get a ring-fence
round his property he bought the four intervening triangular fields. The
puzzle is to discover what was then the whole area of his estate.

191.—THE CRESCENT PUZZLE.

Here is an easy geometrical puzzle. The crescent is formed by two circles,
and C is the centre of the larger circle. The width of the crescent between B
and D is 9 inches, and between E and F 5 inches. What are the diameters of
the two circles?

192.—THE PUZZLE WALL.

There was a small lake, around which four poor men built their cottages.
Four rich men afterwards built their mansions, as shown in the illustration,
and they wished to have the lake to themselves, so they instructed a builder
to put up the shortest possible wall that would exclude the cottagers, but
give themselves free access to the lake. How was the wall to be built?

193.—THE SHEEP-FOLD.
It is a curious fact that the answers always given to some of the best-known
puzzles that appear in every little book of fireside recreations that has been
published for the last fifty or a hundred years are either quite unsatisfactory
or clearly wrong. Yet nobody ever seems to detect their faults. Here is an
example:—A farmer had a pen made of fifty hurdles, capable of holding a
hundred sheep only. Supposing he wanted to make it sufficiently large to
hold double that number, how many additional hurdles must he have?

194.—THE GARDEN WALLS.

A speculative country builder has a circular field, on which he has erected
four cottages, as shown in the illustration. The field is surrounded by a brick
wall, and the owner undertook to put up three other brick walls, so that the
neighbours should not be overlooked by each other, but the four tenants
insist that there shall be no favouritism, and that each shall have exactly the
same length of wall space for his wall fruit trees. The puzzle is to show how
the three walls may be built so that each tenant shall have the same area of
ground, and precisely the same length of wall.
Of course, each garden must be entirely enclosed by its walls, and it must
be possible to prove that each garden has exactly the same length of wall. If
the puzzle is properly solved no figures are necessary.

195.—LADY BELINDA'S GARDEN.

Lady Belinda is an enthusiastic gardener. In the illustration she is depicted
in the act of worrying out a pleasant little problem which I will relate. One
of her gardens is oblong in shape, enclosed by a high holly hedge, and she
is turning it into a rosary for the cultivation of some of her choicest roses.
She wants to devote exactly half of the area of the garden to the flowers, in
one large bed, and the other half to be a path going all round it of equal
breadth throughout. Such a garden is shown in the diagram at the foot of the
picture. How is she to mark out the garden under these simple conditions?
She has only a tape, the length of the garden, to do it with, and, as the holly
hedge is so thick and dense, she must make all her measurements inside.
Lady Belinda did not know the exact dimensions of the garden, and, as it
was not necessary for her to know, I also give no dimensions. It is quite a
simple task no matter what the size or proportions of the garden may be. Yet
how many lady gardeners would know just how to proceed? The tape may
be quite plain—that is, it need not be a graduated measure.

196.—THE TETHERED GOAT.

Here is a little problem that everybody should know how to solve. The goat
is placed in a half-acre meadow, that is in shape an equilateral triangle. It is
tethered to a post at one corner of the field. What should be the length of the
tether (to the nearest inch) in order that the goat shall be able to eat just half
the grass in the field? It is assumed that the goat can feed to the end of the
tether.

197.—THE COMPASSES PUZZLE.
It is curious how an added condition or restriction will sometimes convert
an absurdly easy puzzle into an interesting and perhaps difficult one. I
remember buying in the street many years ago a little mechanical puzzle
that had a tremendous sale at the time. It consisted of a medal with holes in
it, and the puzzle was to work a ring with a gap in it from hole to hole until
it was finally detached. As I was walking along the street I very soon
acquired the trick of taking off the ring with one hand while holding the
puzzle in my pocket. A friend to whom I showed the little feat set about
accomplishing it himself, and when I met him some days afterwards he

exhibited his proficiency in the art. But he was a little taken aback when I
then took the puzzle from him and, while simply holding the medal between
the finger and thumb of one hand, by a series of little shakes and jerks
caused the ring, without my even touching it, to fall off upon the floor. The
following little poser will probably prove a rather tough nut for a great
many readers, simply on account of the restricted conditions:—
Show how to find exactly the middle of any straight line by means of the
compasses only. You are not allowed to use any ruler, pencil, or other article
—only the compasses; and no trick or dodge, such as folding the paper, will
be permitted. You must simply use the compasses in the ordinary legitimate
way.

198.—THE EIGHT STICKS.
I have eight sticks, four of them being exactly half the length of the others. I
lay every one of these on the table, so that they enclose three squares, all of
the same size. How do I do it? There must be no loose ends hanging over.

199.—PAPA'S PUZZLE.
Here is a puzzle by Pappus, who lived at Alexandria about the end of the
third century. It is the fifth proposition in the eighth book of his
Mathematical Collections. I give it in the form that I presented it some
years ago under the title "Papa's Puzzle," just to see how many readers
would discover that it was by Pappus himself. "The little maid's papa has
taken two different-sized rectangular pieces of cardboard, and has clipped
off a triangular piece from one of them, so that when it is suspended by a
thread from the point A it hangs with the long side perfectly horizontal, as
shown in the illustration. He has perplexed the child by asking her to find
the point A on the other card, so as to produce a similar result when cut and
suspended by a thread." Of course, the point must not be found by trial

clippings. A curious and pretty point is involved in this setting of the puzzle.
Can the reader discover it?

200.—A KITE-FLYING PUZZLE.
While accompanying my friend Professor Highflite during a scientific kite-
flying competition on the South Downs of Sussex I was led into a little
calculation that ought to interest my readers. The Professor was paying out
the wire to which his kite was attached from a winch on which it had been
rolled into a perfectly spherical form. This ball of wire was just two feet in
diameter, and the wire had a diameter of one-hundredth of an inch. What
was the length of the wire?
Now, a simple little question like this that everybody can perfectly
understand will puzzle many people to answer in any way. Let us see
whether, without going into any profound mathematical calculations, we

can get the answer roughly—say, within a mile of what is correct! We will
assume that when the wire is all wound up the ball is perfectly solid
throughout, and that no allowance has to be made for the axle that passes
through it. With that simplification, I wonder how many readers can state
within even a mile of the correct answer the length of that wire.

201.—HOW TO MAKE CISTERNS.

Our friend in the illustration has a large sheet of zinc, measuring (before
cutting) eight feet by three feet, and he has cut out square pieces (all of the
same size) from the four corners and now proposes to fold up the sides,
solder the edges, and make a cistern. But the point that puzzles him is this:
Has he cut out those square pieces of the correct size in order that the

cistern may hold the greatest possible quantity of water? You see, if you cut
them very small you get a very shallow cistern; if you cut them large you
get a tall and slender one. It is all a question of finding a way of cutting put
these four square pieces exactly the right size. How are we to avoid making
them too small or too large?

202.—THE CONE PUZZLE.

I have a wooden cone, as shown in Fig. 1. How am I to cut out of it the
greatest possible cylinder? It will be seen that I can cut out one that is long
and slender, like Fig. 2, or short and thick, like Fig. 3. But neither is the
largest possible. A child could tell you where to cut, if he knew the rule.
Can you find this simple rule?

203.—CONCERNING WHEELS.

There are some curious facts concerning the movements of wheels that are
apt to perplex the novice. For example: when a railway train is travelling
from London to Crewe certain parts of the train at any given moment are
actually moving from Crewe towards London. Can you indicate those
parts? It seems absurd that parts of the same train can at any time travel in
opposite directions, but such is the case.
In the accompanying illustration we have two wheels. The lower one is
supposed to be fixed and the upper one running round it in the direction of
the arrows. Now, how many times does the upper wheel turn on its own
axis in making a complete revolution of the other wheel? Do not be in a
hurry with your answer, or you are almost certain to be wrong. Experiment

with two pennies on the table and the correct answer will surprise you,
when you succeed in seeing it.

204.—A NEW MATCH PUZZLE.

In the illustration eighteen matches are shown arranged so that they enclose
two spaces, one just twice as large as the other. Can you rearrange them (1)
so as to enclose two four-sided spaces, one exactly three times as large as
the other, and (2) so as to enclose two five-sided spaces, one exactly three
times as large as the other? All the eighteen matches must be fairly used in
each case; the two spaces must be quite detached, and there must be no
loose ends or duplicated matches.

205.—THE SIX SHEEP-PENS.

Here is a new little puzzle with matches. It will be seen in the illustration
that thirteen matches, representing a farmer's hurdles, have been so placed
that they enclose six sheep-pens all of the same size. Now, one of these
hurdles was stolen, and the farmer wanted still to enclose six pens of equal
size with the remaining twelve. How was he to do it? All the twelve
matches must be fairly used, and there must be no duplicated matches or
loose ends.

POINTS AND LINES PROBLEMS.
"Line upon line, line upon line; here a little and there a little."—Isa. xxviii.
10.
What are known as "Points and Lines" puzzles are found very interesting by
many people. The most familiar example, here given, to plant nine trees so
that they shall form ten straight rows with three trees in every row, is
attributed to Sir Isaac Newton, but the earliest collection of such puzzles is,
I believe, in a rare little book that I possess—published in 1821—Rational
Amusement for Winter Evenings, by John Jackson. The author gives ten
examples of "Trees planted in Rows."

These tree-planting puzzles have always been a matter of great perplexity.
They are real "puzzles," in the truest sense of the word, because nobody has
yet succeeded in finding a direct and certain way of solving them. They
demand the exercise of sagacity, ingenuity, and patience, and what we call
"luck" is also sometimes of service. Perhaps some day a genius will
discover the key to the whole mystery. Remember that the trees must be
regarded as mere points, for if we were allowed to make our trees big
enough we might easily "fudge" our diagrams and get in a few extra straight
rows that were more apparent than real.

206.—THE KING AND THE CASTLES.
There was once, in ancient times, a powerful king, who had eccentric ideas
on the subject of military architecture. He held that there was great strength
and economy in symmetrical forms, and always cited the example of the
bees, who construct their combs in perfect hexagonal cells, to prove that he
had nature to support him. He resolved to build ten new castles in his
country all to be connected by fortified walls, which should form five lines
with four castles in every line. The royal architect presented his preliminary
plan in the form I have shown. But the monarch pointed out that every
castle could be approached from the outside, and commanded that the plan
should be so modified that as many castles as possible should be free from
attack from the outside, and could only be reached by crossing the fortified
walls. The architect replied that he thought it impossible so to arrange them
that even one castle, which the king proposed to use as a royal residence,
could be so protected, but his majesty soon enlightened him by pointing out
how it might be done. How would you have built the ten castles and
fortifications so as best to fulfil the king's requirements? Remember that
they must form five straight lines with four castles in every line.

207.—CHERRIES AND PLUMS.

The illustration is a plan of a cottage as it stands surrounded by an orchard
of fifty-five trees. Ten of these trees are cherries, ten are plums, and the
remainder apples. The cherries are so planted as to form five straight lines,
with four cherry trees in every line. The plum trees are also planted so as to
form five straight lines with four plum trees in every line. The puzzle is to
show which are the ten cherry trees and which are the ten plums. In order
that the cherries and plums should have the most favourable aspect, as few
as possible (under the conditions) are planted on the north and east sides of
the orchard. Of course in picking out a group of ten trees (cherry or plum,
as the case may be) you ignore all intervening trees. That is to say, four
trees may be in a straight line irrespective of other trees (or the house) being
in between. After the last puzzle this will be quite easy.

208.—A PLANTATION PUZZLE.

A man had a square plantation of forty-nine trees, but, as will be seen by the
omissions in the illustration, four trees were blown down and removed. He
now wants to cut down all the remainder except ten trees, which are to be
so left that they shall form five straight rows with four trees in every row.
Which are the ten trees that he must leave?

209.—THE TWENTY-ONE TREES.
A gentleman wished to plant twenty-one trees in his park so that they
should form twelve straight rows with five trees in every row. Could you
have supplied him with a pretty symmetrical arrangement that would satisfy
these conditions?

210.—THE TEN COINS.

Place ten pennies on a large sheet of paper or cardboard, as shown in the
diagram, five on each edge. Now remove four of the coins, without
disturbing the others, and replace them on the paper so that the ten shall
form five straight lines with four coins in every line. This in itself is not
difficult, but you should try to discover in how many different ways the
puzzle may be solved, assuming that in every case the two rows at starting
are exactly the same.

211.—THE TWELVE MINCE-PIES.
It will be seen in our illustration how twelve mince-pies may be placed on
the table so as to form six straight rows with four pies in every row. The
puzzle is to remove only four of them to new positions so that there shall be
seven straight rows with four in every row. Which four would you remove,
and where would you replace them?

212.—THE BURMESE PLANTATION.
A short time ago I received an interesting communication from the British
chaplain at Meiktila, Upper Burma, in which my correspondent informed
me that he had found some amusement on board ship on his way out in
trying to solve this little poser.

If he has a plantation of forty-nine trees, planted in the form of a square as
shown in the accompanying illustration, he wishes to know how he may cut
down twenty-seven of the trees so that the twenty-two left standing shall
form as many rows as possible with four trees in every row.
Of course there may not be more than four trees in any row.

213.—TURKS AND RUSSIANS.
This puzzle is on the lines of the Afridi problem published by me in Tit-Bits
some years ago.
On an open level tract of country a party of Russian infantry, no two of
whom were stationed at the same spot, were suddenly surprised by thirty-
two Turks, who opened fire on the Russians from all directions. Each of the
Turks simultaneously fired a bullet, and each bullet passed immediately
over the heads of three Russian soldiers. As each of these bullets when fired

killed a different man, the puzzle is to discover what is the smallest possible
number of soldiers of which the Russian party could have consisted and
what were the casualties on each side.

MOVING COUNTER PROBLEMS.
"I cannot do't without counters."
Winter's Tale, iv. 3.
Puzzles of this class, except so far as they occur in connection with actual
games, such as chess, seem to be a comparatively modern introduction.
Mathematicians in recent times, notably Vandermonde and Reiss, have
devoted some attention to them, but they do not appear to have been
considered by the old writers. So far as games with counters are concerned,
perhaps the most ancient and widely known in old times is "Nine Men's
Morris" (known also, as I shall show, under a great many other names),
unless the simpler game, distinctly mentioned in the works of Ovid (No.
110, "Ovid's Game," in The Canterbury Puzzles), from which "Noughts and
Crosses" seems to be derived, is still more ancient.
In France the game is called Marelle, in Poland Siegen Wulf Myll (She-goat
Wolf Mill, or Fight), in Germany and Austria it is called Muhle (the Mill),
in Iceland it goes by the name of Mylla, while the Bogas (or native bargees)
of South America are said to play it, and on the Amazon it is called Trique,
and held to be of Indian origin. In our own country it has different names in
different districts, such as Meg Merrylegs, Peg Meryll, Nine Peg o'Merryal,
Nine-Pin Miracle, Merry Peg, and Merry Hole. Shakespeare refers to it in
"Midsummer Night's Dream" (Act ii., scene 1):—
"The nine-men's morris is filled up with mud;
And the quaint mazes in the wanton green,
For lack of tread, are undistinguishable."
It was played by the shepherds with stones in holes cut in the turf. John
Clare, the peasant poet of Northamptonshire, in "The Shepherd Boy" (1835)
says:—"Oft we track his haunts .... By nine-peg-morris nicked upon the
green." It is also mentioned by Drayton in his "Polyolbion."
It was found on an old Roman tile discovered during the excavations at
Silchester, and cut upon the steps of the Acropolis at Athens. When visiting

the Christiania Museum a few years ago I was shown the great Viking ship
that was discovered at Gokstad in 1880. On the oak planks forming the
deck of the vessel were found boles and lines marking out the game, the
holes being made to receive pegs. While inspecting the ancient oak
furniture in the Rijks Museum at Amsterdam I became interested in an old
catechumen's settle, and was surprised to find the game diagram cut in the
centre of the seat—quite conveniently for surreptitious play. It has been
discovered cut in the choir stalls of several of our English cathedrals. In the
early eighties it was found scratched upon a stone built into a wall
(probably about the date 1200), during the restoration of Hargrave church in
Northamptonshire. This stone is now in the Northampton Museum. A
similar stone has since been found at Sempringham, Lincolnshire. It is to be
seen on an ancient tombstone in the Isle of Man, and painted on old Dutch
tiles. And in 1901 a stone was dug out of a gravel pit near Oswestry bearing
an undoubted diagram of the game.
The game has been played with different rules at different periods and
places. I give a copy of the board. Sometimes the diagonal lines are
omitted, but this evidently was not intended to affect the play: it simply
meant that the angles alone were thought sufficient to indicate the points.
This is how Strutt, in Sports and Pastimes, describes the game, and it agrees
with the way I played it as a boy:—"Two persons, having each of them nine
pieces, or men, lay them down alternately, one by one, upon the spots; and
the business of either party is to prevent his antagonist from placing three of
his pieces so as to form a row of three, without the intervention of an
opponent piece. If a row be formed, he that made it is at liberty to take up
one of his competitor's pieces from any part he thinks most to his
advantage; excepting he has made a row, which must not be touched if he
have another piece upon the board that is not a component part of that row.
When all the pieces are laid down, they are played backwards and forwards,
in any direction that the lines run, but only can move from one spot to
another (next to it) at one time. He that takes off all his antagonist's pieces
is the conqueror."

214.—THE SIX FROGS.

The six educated frogs in the illustration are trained to reverse their order,
so that their numbers shall read 6, 5, 4, 3, 2, 1, with the blank square in its
present position. They can jump to the next square (if vacant) or leap over
one frog to the next square beyond (if vacant), just as we move in the game
of draughts, and can go backwards or forwards at pleasure. Can you show
how they perform their feat in the fewest possible moves? It is quite easy,
so when you have done it add a seventh frog to the right and try again. Then
add more frogs until you are able to give the shortest solution for any
number. For it can always be done, with that single vacant square, no matter
how many frogs there are.

215.—THE GRASSHOPPER PUZZLE.
It has been suggested that this puzzle was a great favourite among the
young apprentices of the City of London in the sixteenth and seventeenth
centuries. Readers will have noticed the curious brass grasshopper on the
Royal Exchange. This long-lived creature escaped the fires of 1666 and
1838. The grasshopper, after his kind, was the crest of Sir Thomas
Gresham, merchant grocer, who died in 1579, and from this cause it has
been used as a sign by grocers in general. Unfortunately for the legend as to
its origin, the puzzle was only produced by myself so late as the year 1900.
On twelve of the thirteen black discs are placed numbered counters or
grasshoppers. The puzzle is to reverse their order, so that they shall read, 1,
2, 3, 4, etc., in the opposite direction, with the vacant disc left in the same
position as at present. Move one at a time in any order, either to the
adjoining vacant disc or by jumping over one grasshopper, like the moves
in draughts. The moves or leaps may be made in either direction that is at
any time possible. What are the fewest possible moves in which it can be
done?

216.—THE EDUCATED FROGS.

Our six educated frogs have learnt a new and pretty feat. When placed on
glass tumblers, as shown in the illustration, they change sides so that the
three black ones are to the left and the white frogs to the right, with the
unoccupied tumbler at the opposite end—No. 7. They can jump to the next
tumbler (if unoccupied), or over one, or two, frogs to an unoccupied
tumbler. The jumps can be made in either direction, and a frog may jump
over his own or the opposite colour, or both colours. Four suecessive
specimen jumps will make everything quite plain: 4 to 1, 5 to 4, 3 to 5, 6 to
3. Can you show how they do it in ten jumps?

217.—THE TWICKENHAM PUZZLE.

In the illustration we have eleven discs in a circle. On five of the discs we
place white counters with black letters—as shown—and on five other discs
the black counters with white letters. The bottom disc is left vacant. Starting
thus, it is required to get the counters into order so that they spell the word
"Twickenham" in a clockwise direction, leaving the vacant disc in the
original position. The black counters move in the direction that a clock-
hand revolves, and the white counters go the opposite way. A counter may
jump over one of the opposite colour if the vacant disc is next beyond.
Thus, if your first move is with K, then C can jump over K. If then K moves
towards E, you may next jump W over C, and so on. The puzzle may be
solved in twenty-six moves. Remember a counter cannot jump over one of
its own colour.

218.—THE VICTORIA CROSS PUZZLE.

The puzzle-maker is peculiarly a "snapper-up of unconsidered trifles," and
his productions are often built up with the slenderest materials. Trivialities
that might entirely escape the observation of others, or, if they were
observed, would be regarded as of no possible moment, often supply the
man who is in quest of posers with a pretty theme or an idea that he thinks
possesses some "basal value."
When seated opposite to a lady in a railway carriage at the time of Queen
Victoria's Diamond Jubilee, my attention was attracted to a brooch that she
was wearing. It was in the form of a Maltese or Victoria Cross, and bore the
letters of the word VICTORIA. The number and arrangement of the letters
immediately gave me the suggestion for the puzzle which I now present.
The diagram, it will be seen, is composed of nine divisions. The puzzle is to
place eight counters, bearing the letters of the word VICTORIA, exactly in
the manner shown, and then slide one letter at a time from black to white
and white to black alternately, until the word reads round in the same
direction, only with the initial letter V on one of the black arms of the cross.

At no time may two letters be in the same division. It is required to find the
shortest method.
Leaping moves are, of course, not permitted. The first move must obviously
be made with A, I, T, or R. Supposing you move T to the centre, the next
counter played will be O or C, since I or R cannot be moved. There is
something a little remarkable in the solution of this puzzle which I will
explain.

219.—THE LETTER BLOCK PUZZLE.

Here is a little reminiscence of our old friend the Fifteen Block Puzzle.
Eight wooden blocks are lettered, and are placed in a box, as shown in the
illustration. It will be seen that you can only move one block at a time to the
place vacant for the time being, as no block may be lifted out of the box.
The puzzle is to shift them about until you get them in the order—

A B C
D E F
G H
This you will find by no means difficult if you are allowed as many moves
as you like. But the puzzle is to do it in the fewest possible moves. I will not
say what this smallest number of moves is, because the reader may like to
discover it for himself. In writing down your moves you will find it
necessary to record no more than the letters in the order that they are
shifted. Thus, your first five moves might be C, H, G, E, F; and this
notation can have no possible ambiguity. In practice you only need eight
counters and a simple diagram on a sheet of paper.

220.—A LODGING-HOUSE DIFFICULTY.

The Dobsons secured apartments at Slocomb-on-Sea. There were six rooms
on the same floor, all communicating, as shown in the diagram. The rooms
they took were numbers 4, 5, and 6, all facing the sea. But a little difficulty
arose. Mr. Dobson insisted that the piano and the bookcase should change
rooms. This was wily, for the Dobsons were not musical, but they wanted to
prevent any one else playing the instrument. Now, the rooms were very
small and the pieces of furniture indicated were very big, so that no two of
these articles could be got into any room at the same time. How was the

exchange to be made with the least possible labour? Suppose, for example,
you first move the wardrobe into No. 2; then you can move the bookcase to
No. 5 and the piano to No. 6, and so on. It is a fascinating puzzle, but the
landlady had reasons for not appreciating it. Try to solve her difficulty in
the fewest possible removals with counters on a sheet of paper.

221.—THE EIGHT ENGINES.
The diagram represents the engine-yard of a railway company under
eccentric management. The engines are allowed to be stationary only at the
nine points indicated, one of which is at present vacant. It is required to
move the engines, one at a time, from point to point, in seventeen moves, so
that their numbers shall be in numerical order round the circle, with the
central point left vacant. But one of the engines has had its fire drawn, and
therefore cannot move. How is the thing to be done? And which engine
remains stationary throughout?

222.—A RAILWAY PUZZLE.

Make a diagram, on a large sheet of paper, like the illustration, and have
three counters marked A, three marked B, and three marked C. It will be
seen that at the intersection of lines there are nine stopping-places, and a
tenth stopping-place is attached to the outer circle like the tail of a Q. Place
the three counters or engines marked A, the three marked B, and the three
marked C at the places indicated. The puzzle is to move the engines, one at
a time, along the lines, from stopping-place to stopping-place, until you
succeed in getting an A, a B, and a C on each circle, and also A, B, and C
on each straight line. You are required to do this in as few moves as
possible. How many moves do you need?

223.—A RAILWAY MUDDLE.

The plan represents a portion of the line of the London, Clodville, and
Mudford Railway Company. It is a single line with a loop. There is only
room for eight wagons, or seven wagons and an engine, between B and C
on either the left line or the right line of the loop. It happened that two
goods trains (each consisting of an engine and sixteen wagons) got into the
position shown in the illustration. It looked like a hopeless deadlock, and
each engine-driver wanted the other to go back to the next station and take
off nine wagons. But an ingenious stoker undertook to pass the trains and
send them on their respective journeys with their engines properly in front.
He also contrived to reverse the engines the fewest times possible. Could
you have performed the feat? And how many times would you require to
reverse the engines? A "reversal" means a change of direction, backward or
forward. No rope-shunting, fly-shunting, or other trick is allowed. All the
work must be done legitimately by the two engines. It is a simple but
interesting puzzle if attempted with counters.

224.—THE MOTOR-GARAGE PUZZLE.

The difficulties of the proprietor of a motor garage are converted into a little
pastime of a kind that has a peculiar fascination. All you need is to make a
simple plan or diagram on a sheet of paper or cardboard and number eight
counters, 1 to 8. Then a whole family can enter into an amusing competition
to find the best possible solution of the difficulty.
The illustration represents the plan of a motor garage, with accommodation
for twelve cars. But the premises are so inconveniently restricted that the
proprietor is often caused considerable perplexity. Suppose, for example,
that the eight cars numbered 1 to 8 are in the positions shown, how are they
to be shifted in the quickest possible way so that 1, 2, 3, and 4 shall change
places with 5, 6, 7, and 8—that is, with the numbers still running from left
to right, as at present, but the top row exchanged with the bottom row?
What are the fewest possible moves?

One car moves at a time, and any distance counts as one move. To prevent
misunderstanding, the stopping-places are marked in squares, and only one
car can be in a square at the same time.

225.—THE TEN PRISONERS.
If prisons had no other use, they might still be preserved for the special
benefit of puzzle-makers. They appear to be an inexhaustible mine of
perplexing ideas. Here is a little poser that will perhaps interest the reader
for a short period. We have in the illustration a prison of sixteen cells. The
locations of the ten prisoners will be seen. The jailer has queer superstitions
about odd and even numbers, and he wants to rearrange the ten prisoners so
that there shall be as many even rows of men, vertically, horizontally, and
diagonally, as possible. At present it will be seen, as indicated by the
arrows, that there are only twelve such rows of 2 and 4. I will state at once
that the greatest number of such rows that is possible is sixteen. But the
jailer only allows four men to be removed to other cells, and informs me
that, as the man who is seated in the bottom right-hand corner is infirm, he
must not be moved. Now, how are we to get those sixteen rows of even
numbers under such conditions?

226.—ROUND THE COAST.

Here is a puzzle that will, I think, be found as amusing as instructive. We
are given a ring of eight circles. Leaving circle 8 blank, we are required to
write in the name of a seven-lettered port in the United Kingdom in this
manner. Touch a blank circle with your pencil, then jump over two circles
in either direction round the ring, and write down the first letter. Then touch
another vacant circle, jump over two circles, and write down your second
letter. Proceed similarly with the other letters in their proper order until you
have completed the word. Thus, suppose we select "Glasgow," and proceed
as follows: 6—1, 7—2, 8—3, 7—4, 8—5, which means that we touch 6,
jump over 7 and and write down "G" on 1; then touch 7, jump over 8 and 1,
and write down "l" on 2; and so on. It will be found that after we have
written down the first five letters—"Glasg"—as above, we cannot go any
further. Either there is something wrong with "Glasgow," or we have not
managed our jumps properly. Can you get to the bottom of the mystery?

227.—CENTRAL SOLITAIRE.

This ancient puzzle was a great favourite with our grandmothers, and most
of us, I imagine, have on occasions come across a "Solitaire" board—a
round polished board with holes cut in it in a geometrical pattern, and a
glass marble in every hole. Sometimes I have noticed one on a side table in
a suburban front parlour, or found one on a shelf in a country cottage, or
had one brought under my notice at a wayside inn. Sometimes they are of
the form shown above, but it is equally common for the board to have four
more holes, at the points indicated by dots. I select the simpler form.
Though "Solitaire" boards are still sold at the toy shops, it will be sufficient
if the reader will make an enlarged copy of the above on a sheet of
cardboard or paper, number the "holes," and provide himself with 33
counters, buttons, or beans. Now place a counter in every hole except the
central one, No. 17, and the puzzle is to take off all the counters in a series
of jumps, except the last counter, which must be left in that central hole.
You are allowed to jump one counter over the next one to a vacant hole
beyond, just as in the game of draughts, and the counter jumped over is
immediately taken off the board. Only remember every move must be a
jump; consequently you will take off a counter at each move, and thirty-one

single jumps will of course remove all the thirty-one counters. But
compound moves are allowed (as in draughts, again), for so long as one
counter continues to jump, the jumps all count as one move.
Here is the beginning of an imaginary solution which will serve to make the
manner of moving perfectly plain, and show how the solver should write
out his attempts: 5-17, 12-10, 26-12, 24-26 (13-11, 11-25), 9-11 (26-24, 24-
10, 10-12), etc., etc. The jumps contained within brackets count as one
move, because they are made with the same counter. Find the fewest
possible moves. Of course, no diagonal jumps are permitted; you can only
jump in the direction of the lines.

228.—THE TEN APPLES.

The family represented in the illustration are amusing themselves with this
little puzzle, which is not very difficult but quite interesting. They have, it

will be seen, placed sixteen plates on the table in the form of a square, and
put an apple in each of ten plates. They want to find a way of removing all
the apples except one by jumping over one at a time to the next vacant
square, as in draughts; or, better, as in solitaire, for you are not allowed to
make any diagonal moves—only moves parallel to the sides of the square.
It is obvious that as the apples stand no move can be made, but you are
permitted to transfer any single apple you like to a vacant plate before
starting. Then the moves must be all leaps, taking off the apples leaped
over.

229.—THE NINE ALMONDS.
"Here is a little puzzle," said a Parson, "that I have found peculiarly
fascinating. It is so simple, and yet it keeps you interested indefinitely."
The reverend gentleman took a sheet of paper and divided it off into
twenty-five squares, like a square portion of a chessboard. Then he placed
nine almonds on the central squares, as shown in the illustration, where we
have represented numbered counters for convenience in giving the solution.
"Now, the puzzle is," continued the Parson, "to remove eight of the almonds
and leave the ninth in the central square. You make the removals by
jumping one almond over another to the vacant square beyond and taking
off the one jumped over—just as in draughts, only here you can jump in any
direction, and not diagonally only. The point is to do the thing in the fewest
possible moves."
The following specimen attempt will make everything clear. Jump 4 over 1,
5 over 9, 3 over 6, 5 over 3, 7 over 5 and 2, 4 over 7, 8 over 4. But 8 is not
left in the central square, as it should be. Remember to remove those you
jump over. Any number of jumps in succession with the same almond count
as one move.

230.—THE TWELVE PENNIES.
Here is a pretty little puzzle that only requires twelve pennies or counters.
Arrange them in a circle, as shown in the illustration. Now take up one
penny at a time and, passing it over two pennies, place it on the third penny.
Then take up another single penny and do the same thing, and so on, until,
in six such moves, you have the coins in six pairs in the positions 1, 2, 3, 4,
5, 6. You can move in either direction round the circle at every play, and it
does not matter whether the two jumped over are separate or a pair. This is
quite easy if you use just a little thought.

231.—PLATES AND COINS.
Place twelve plates, as shown, on a round table, with a penny or orange in
every plate. Start from any plate you like and, always going in one direction
round the table, take up one penny, pass it over two other pennies, and place
it in the next plate. Go on again; take up another penny and, having passed
it over two pennies, place it in a plate; and so continue your journey. Six
coins only are to be removed, and when these have been placed there
should be two coins in each of six plates and six plates empty. An important
point of the puzzle is to go round the table as few times as possible. It does
not matter whether the two coins passed over are in one or two plates, nor
how many empty plates you pass a coin over. But you must always go in
one direction round the table and end at the point from which you set out.
Your hand, that is to say, goes steadily forward in one direction, without
ever moving backwards.

232.—CATCHING THE MICE.

"Play fair!" said the mice. "You know the rules of the game."
"Yes, I know the rules," said the cat. "I've got to go round and round the
circle, in the direction that you are looking, and eat every thirteenth mouse,
but I must keep the white mouse for a tit-bit at the finish. Thirteen is an
unlucky number, but I will do my best to oblige you."
"Hurry up, then!" shouted the mice.
"Give a fellow time to think," said the cat. "I don't know which of you to
start at. I must figure it out."
While the cat was working out the puzzle he fell asleep, and, the spell being
thus broken, the mice returned home in safety. At which mouse should the
cat have started the count in order that the white mouse should be the last
eaten?
When the reader has solved that little puzzle, here is a second one for him.
What is the smallest number that the cat can count round and round the
circle, if he must start at the white mouse (calling that "one" in the count)
and still eat the white mouse last of all?
And as a third puzzle try to discover what is the smallest number that the
cat can count round and round if she must start at the white mouse (calling
that "one") and make the white mouse the third eaten.

233.—THE ECCENTRIC CHEESEMONGER.

The cheesemonger depicted in the illustration is an inveterate puzzle lover.
One of his favourite puzzles is the piling of cheeses in his warehouse, an
amusement that he finds good exercise for the body as well as for the mind.
He places sixteen cheeses on the floor in a straight row and then makes
them into four piles, with four cheeses in every pile, by always passing a
cheese over four others. If you use sixteen counters and number them in
order from 1 to 16, then you may place 1 on 6, 11 on 1, 7 on 4, and so on,
until there are four in every pile. It will be seen that it does not matter
whether the four passed over are standing alone or piled; they count just the
same, and you can always carry a cheese in either direction. There are a
great many different ways of doing it in twelve moves, so it makes a good
game of "patience" to try to solve it so that the four piles shall be left in
different stipulated places. For example, try to leave the piles at the extreme
ends of the row, on Nos. 1, 2, 15 and 16; this is quite easy. Then try to leave

three piles together, on Nos. 13, 14, and 15. Then again play so that they
shall be left on Nos. 3, 5, 12, and 14.

234.—THE EXCHANGE PUZZLE.
Here is a rather entertaining little puzzle with moving counters. You only
need twelve counters—six of one colour, marked A, C, E, G, I, and K, and
the other six marked B, D, F, H, J, and L. You first place them on the
diagram, as shown in the illustration, and the puzzle is to get them into
regular alphabetical order, as follows:—

ABCD
E F GH
I J KL

The moves are made by exchanges of opposite colours standing on the
same line. Thus, G and J may exchange places, or F and A, but you cannot
exchange G and C, or F and D, because in one case they are both white and
in the other case both black. Can you bring about the required arrangement
in seventeen exchanges?

It cannot be done in fewer moves. The puzzle is really much easier than it
looks, if properly attacked.

235.—TORPEDO PRACTICE.

If a fleet of sixteen men-of-war were lying at anchor and surrounded by the
enemy, how many ships might be sunk if every torpedo, projected in a
straight line, passed under three vessels and sank the fourth? In the diagram
we have arranged the fleet in square formation, where it will be seen that as
many as seven ships may be sunk (those in the top row and first column) by
firing the torpedoes indicated by arrows. Anchoring the fleet as we like, to
what extent can we increase this number? Remember that each successive
ship is sunk before another torpedo is launched, and that every torpedo
proceeds in a different direction; otherwise, by placing the ships in a
straight line, we might sink as many as thirteen! It is an interesting little
study in naval warfare, and eminently practical—provided the enemy will

allow you to arrange his fleet for your convenience and promise to lie still
and do nothing!

236.—THE HAT PUZZLE.
Ten hats were hung on pegs as shown in the illustration—five silk hats and
five felt "bowlers," alternately silk and felt. The two pegs at the end of the
row were empty.

The puzzle is to remove two contiguous hats to the vacant pegs, then two
other adjoining hats to the pegs now unoccupied, and so on until five pairs
have been moved and the hats again hang in an unbroken row, but with all
the silk ones together and all the felt hats together.
Remember, the two hats removed must always be contiguous ones, and you
must take one in each hand and place them on their new pegs without
reversing their relative position. You are not allowed to cross your hands,
nor to hang up one at a time.
Can you solve this old puzzle, which I give as introductory to the next? Try
it with counters of two colours or with coins, and remember that the two
empty pegs must be left at one end of the row.

237.—BOYS AND GIRLS.
If you mark off ten divisions on a sheet of paper to represent the chairs, and
use eight numbered counters for the children, you will have a fascinating
pastime. Let the odd numbers represent boys and even numbers girls, or
you can use counters of two colours, or coins.

The puzzle is to remove two children who are occupying adjoining chairs
and place them in two empty chairs, making them first change sides; then
remove a second pair of children from adjoining chairs and place them in
the two now vacant, making them change sides; and so on, until all the boys
are together and all the girls together, with the two vacant chairs at one end
as at present. To solve the puzzle you must do this in five moves. The two
children must always be taken from chairs that are next to one another; and
remember the important point of making the two children change sides, as
this latter is the distinctive feature of the puzzle. By "change sides" I simply
mean that if, for example, you first move 1 and 2 to the vacant chairs, then
the first (the outside) chair will be occupied by 2 and the second one by 1.

238.—ARRANGING THE JAMPOTS.
I happened to see a little girl sorting out some jam in a cupboard for her
mother. She was putting each different kind of preserve apart on the
shelves. I noticed that she took a pot of damson in one hand and a pot of
gooseberry in the other and made them change places; then she changed a
strawberry with a raspberry, and so on. It was interesting to observe what a
lot of unnecessary trouble she gave herself by making more interchanges
than there was any need for, and I thought it would work into a good puzzle.

It will be seen in the illustration that little Dorothy has to manipulate
twenty-four large jampots in as many pigeon-holes. She wants to get them
in correct numerical order—that is, 1, 2, 3, 4, 5, 6 on the top shelf, 7, 8, 9,
10, 11, 12 on the next shelf, and so on. Now, if she always takes one pot in
the right hand and another in the left and makes them change places, how
many of these interchanges will be necessary to get all the jampots in
proper order? She would naturally first change the 1 and the 3, then the 2
and the 3, when she would have the first three pots in their places. How
would you advise her to go on then? Place some numbered counters on a
sheet of paper divided into squares for the pigeon-holes, and you will find it
an amusing puzzle.

UNICURSAL AND ROUTE PROBLEMS.
"I see them on their winding way."
REGINALD HEBER.
It is reasonable to suppose that from the earliest ages one man has asked
another such questions as these: "Which is the nearest way home?" "Which
is the easiest or pleasantest way?" "How can we find a way that will enable
us to dodge the mastodon and the plesiosaurus?" "How can we get there
without ever crossing the track of the enemy?" All these are elementary
route problems, and they can be turned into good puzzles by the
introduction of some conditions that complicate matters. A variety of such
complications will be found in the following examples. I have also included
some enumerations of more or less difficulty. These afford excellent
practice for the reasoning faculties, and enable one to generalize in the case
of symmetrical forms in a manner that is most instructive.

239.—A JUVENILE PUZZLE.
For years I have been perpetually consulted by my juvenile friends about
this little puzzle. Most children seem to know it, and yet, curiously enough,
they are invariably unacquainted with the answer. The question they always
ask is, "Do, please, tell me whether it is really possible." I believe Houdin
the conjurer used to be very fond of giving it to his child friends, but I
cannot say whether he invented the little puzzle or not. No doubt a large
number of my readers will be glad to have the mystery of the solution
cleared up, so I make no apology for introducing this old "teaser."
The puzzle is to draw with three strokes of the pencil the diagram that the
little girl is exhibiting in the illustration. Of course, you must not remove
your pencil from the paper during a stroke or go over the same line a second
time. You will find that you can get in a good deal of the figure with one
continuous stroke, but it will always appear as if four strokes are necessary.

Another form of the puzzle is to draw the diagram on a slate and then rub it
out in three rubs.

240.—THE UNION JACK.

The illustration is a rough sketch somewhat resembling the British flag, the
Union Jack. It is not possible to draw the whole of it without lifting the
pencil from the paper or going over the same line twice. The puzzle is to
find out just how much of the drawing it is possible to make without lifting
your pencil or going twice over the same line. Take your pencil and see
what is the best you can do.

241.—THE DISSECTED CIRCLE.
How many continuous strokes, without lifting your pencil from the paper,
do you require to draw the design shown in our illustration? Directly you
change the direction of your pencil it begins a new stroke. You may go over
the same line more than once if you like. It requires just a little care, or you
may find yourself beaten by one stroke.

242.—THE TUBE INSPECTOR'S PUZZLE.
The man in our illustration is in a little dilemma. He has just been appointed
inspector of a certain system of tube railways, and it is his duty to inspect
regularly, within a stated period, all the company's seventeen lines
connecting twelve stations, as shown on the big poster plan that he is
contemplating. Now he wants to arrange his route so that it shall take him
over all the lines with as little travelling as possible. He may begin where he
likes and end where he likes. What is his shortest route?
Could anything be simpler? But the reader will soon find that, however he
decides to proceed, the inspector must go over some of the lines more than
once. In other words, if we say that the stations are a mile apart, he will
have to travel more than seventeen miles to inspect every line. There is the
little difficulty. How far is he compelled to travel, and which route do you
recommend?

243.—VISITING THE TOWNS.

A traveller, starting from town No. 1, wishes to visit every one of the towns
once, and once only, going only by roads indicated by straight lines. How
many different routes are there from which he can select? Of course, he
must end his journey at No. 1, from which he started, and must take no
notice of cross roads, but go straight from town to town. This is an absurdly
easy puzzle, if you go the right way to work.

244.—THE FIFTEEN TURNINGS.
Here is another queer travelling puzzle, the solution of which calls for
ingenuity. In this case the traveller starts from the black town and wishes to
go as far as possible while making only fifteen turnings and never going
along the same road twice. The towns are supposed to be a mile apart.
Supposing, for example, that he went straight to A, then straight to B, then
to C, D, E, and F, you will then find that he has travelled thirty-seven miles
in five turnings. Now, how far can he go in fifteen turnings?

245.—THE FLY ON THE OCTAHEDRON.
"Look here," said the professor to his colleague, "I have been watching that
fly on the octahedron, and it confines its walks entirely to the edges. What
can be its reason for avoiding the sides?"
"Perhaps it is trying to solve some route problem," suggested the other.
"Supposing it to start from the top point, how many different routes are
there by which it may walk over all the edges, without ever going twice
along the same edge in any route?"

The problem was a harder one than they expected, and after working at it
during leisure moments for several days their results did not agree—in fact,
they were both wrong. If the reader is surprised at their failure, let him
attempt the little puzzle himself. I will just explain that the octahedron is
one of the five regular, or Platonic, bodies, and is contained under eight
equal and equilateral triangles. If you cut out the two pieces of cardboard of
the shape shown in the margin of the illustration, cut half through along the
dotted lines and then bend them and put them together, you will have a
perfect octahedron. In any route over all the edges it will be found that the
fly must end at the point of departure at the top.

246.—THE ICOSAHEDRON PUZZLE.
The icosahedron is another of the five regular, or Platonic, bodies having all
their sides, angles, and planes similar and equal. It is bounded by twenty
similar equilateral triangles. If you cut out a piece of cardboard of the form
shown in the smaller diagram, and cut half through along the dotted lines, it
will fold up and form a perfect icosahedron.
Now, a Platonic body does not mean a heavenly body; but it will suit the
purpose of our puzzle if we suppose there to be a habitable planet of this
shape. We will also suppose that, owing to a superfluity of water, the only
dry land is along the edges, and that the inhabitants have no knowledge of
navigation. If every one of those edges is 10,000 miles long and a solitary
traveller is placed at the North Pole (the highest point shown), how far will
he have to travel before he will have visited every habitable part of the
planet—that is, have traversed every one of the edges?

247.—INSPECTING A MINE.
The diagram is supposed to represent the passages or galleries in a mine.
We will assume that every passage, A to B, B to C, C to H, H to I, and so
on, is one furlong in length. It will be seen that there are thirty-one of these

passages. Now, an official has to inspect all of them, and he descends by the
shaft to the point A. How far must he travel, and what route do you
recommend? The reader may at first say, "As there are thirty-one passages,
each a furlong in length, he will have to travel just thirty-one furlongs." But
this is assuming that he need never go along a passage more than once,
which is not the case. Take your pencil and try to find the shortest route.
You will soon discover that there is room for considerable judgment. In
fact, it is a perplexing puzzle.

248.—THE CYCLISTS' TOUR.
Two cyclists were consulting a road map in preparation for a little tour
together. The circles represent towns, and all the good roads are represented
by lines. They are starting from the town with a star, and must complete
their tour at E. But before arriving there they want to visit every other town
once, and only once. That is the difficulty. Mr. Spicer said, "I am certain we
can find a way of doing it;" but Mr. Maggs replied, "No way, I'm sure."
Now, which of them was correct? Take your pencil and see if you can find
any way of doing it. Of course you must keep to the roads indicated.

249.—THE SAILOR'S PUZZLE.
The sailor depicted in the illustration stated that he had since his boyhood
been engaged in trading with a small vessel among some twenty little
islands in the Pacific. He supplied the rough chart of which I have given a
copy, and explained that the lines from island to island represented the only
routes that he ever adopted. He always started from island A at the
beginning of the season, and then visited every island once, and once only,
finishing up his tour at the starting-point A. But he always put off his visit
to C as long as possible, for trade reasons that I need not enter into. The
puzzle is to discover his exact route, and this can be done with certainty.
Take your pencil and, starting at A, try to trace it out. If you write down the
islands in the order in which you visit them—thus, for example, A, I, O, L,
G, etc.—you can at once see if you have visited an island twice or omitted
any. Of course, the crossings of the lines must be ignored—that is, you must
continue your route direct, and you are not allowed to switch off at a
crossing and proceed in another direction. There is no trick of this kind in
the puzzle. The sailor knew the best route. Can you find it?

250.—THE GRAND TOUR.
One of the everyday puzzles of life is the working out of routes. If you are
taking a holiday on your bicycle, or a motor tour, there always arises the
question of how you are to make the best of your time and other resources.
You have determined to get as far as some particular place, to include visits
to such-and-such a town, to try to see something of special interest
elsewhere, and perhaps to try to look up an old friend at a spot that will not
take you much out of your way. Then you have to plan your route so as to
avoid bad roads, uninteresting country, and, if possible, the necessity of a
return by the same way that you went. With a map before you, the
interesting puzzle is attacked and solved. I will present a little poser based
on these lines.
I give a rough map of a country—it is not necessary to say what particular
country—the circles representing towns and the dotted lines the railways
connecting them. Now there lived in the town marked A a man who was
born there, and during the whole of his life had never once left his native
place. From his youth upwards he had been very industrious, sticking
incessantly to his trade, and had no desire whatever to roam abroad.
However, on attaining his fiftieth birthday he decided to see something of
his country, and especially to pay a visit to a very old friend living at the
town marked Z. What he proposed was this: that he would start from his
home, enter every town once and only once, and finish his journey at Z. As
he made up his mind to perform this grand tour by rail only, he found it
rather a puzzle to work out his route, but he at length succeeded in doing so.
How did he manage it? Do not forget that every town has to be visited once,
and not more than once.

251.—WATER, GAS, AND ELECTRICITY.
There are some half-dozen puzzles, as old as the hills, that are perpetually
cropping up, and there is hardly a month in the year that does not bring
inquiries as to their solution. Occasionally one of these, that one had
thought was an extinct volcano, bursts into eruption in a surprising manner.
I have received an extraordinary number of letters respecting the ancient
puzzle that I have called "Water, Gas, and Electricity." It is much older than
electric lighting, or even gas, but the new dress brings it up to date. The
puzzle is to lay on water, gas, and electricity, from W, G, and E, to each of
the three houses, A, B, and C, without any pipe crossing another. Take your
pencil and draw lines showing how this should be done. You will soon find
yourself landed in difficulties.

252.—A PUZZLE FOR MOTORISTS.

Eight motorists drove to church one morning. Their respective houses and
churches, together with the only roads available (the dotted lines), are
shown. One went from his house A to his church A, another from his house
B to his church B, another from C to C, and so on, but it was afterwards
found that no driver ever crossed the track of another car. Take your pencil
and try to trace out their various routes.

253.—A BANK HOLIDAY PUZZLE.
Two friends were spending their bank holiday on a cycling trip. Stopping
for a rest at a village inn, they consulted a route map, which is represented
in our illustration in an exceedingly simplified form, for the puzzle is
interesting enough without all the original complexities. They started from
the town in the top left-hand corner marked A. It will be seen that there are
one hundred and twenty such towns, all connected by straight roads. Now
they discovered that there are exactly 1,365 different routes by which they
may reach their destination, always travelling either due south or due east.
The puzzle is to discover which town is their destination.

Of course, if you find that there are more than 1,365 different routes to a
town it cannot be the right one.

254.—THE MOTOR-CAR TOUR.

In the above diagram the circles represent towns and the lines good roads.
In just how many different ways can a motorist, starting from London
(marked with an L), make a tour of all these towns, visiting every town
once, and only once, on a tour, and always coming back to London on the
last ride? The exact reverse of any route is not counted as different.

255.—THE LEVEL PUZZLE.

This is a simple counting puzzle. In how many different ways can you spell
out the word LEVEL by placing the point of your pencil on an L and then
passing along the lines from letter to letter. You may go in any direction,
backwards or forwards. Of course you are not allowed to miss letters—that
is to say, if you come to a letter you must use it.

256.—THE DIAMOND PUZZLE.
IN how many different ways may the word DIAMOND be read in the
arrangement shown? You may start wherever you like at a D and go up or
down, backwards or forwards, in and out, in any direction you like, so long
as you always pass from one letter to another that adjoins it. How many
ways are there?

257.—THE DEIFIED PUZZLE.
In how many different ways may the word DEIFIED be read in this
arrangement under the same conditions as in the last puzzle, with the
addition that you can use any letters twice in the same reading?

258.—THE VOTERS' PUZZLE.

Here we have, perhaps, the most interesting form of the puzzle. In how
many different ways can you read the political injunction, "RISE TO
VOTE, SIR," under the same conditions as before? In this case every
reading of the palindrome requires the use of the central V as the middle
letter.

259.—HANNAH'S PUZZLE.
A man was in love with a young lady whose Christian name was Hannah.
When he asked her to be his wife she wrote down the letters of her name in
this manner:—

and promised that she would be his if he could tell her correctly in how
many different ways it was possible to spell out her name, always passing
from one letter to another that was adjacent. Diagonal steps are here
allowed. Whether she did this merely to tease him or to test his cleverness is
not recorded, but it is satisfactory to know that he succeeded. Would you
have been equally successful? Take your pencil and try. You may start from
any of the H's and go backwards or forwards and in any direction, so long
as all the letters in a spelling are adjoining one another. How many ways are
there, no two exactly alike?

260.—THE HONEYCOMB PUZZLE.

Here is a little puzzle with the simplest possible conditions. Place the point
of your pencil on a letter in one of the cells of the honeycomb, and trace out
a very familiar proverb by passing always from a cell to one that is
contiguous to it. If you take the right route you will have visited every cell
once, and only once. The puzzle is much easier than it looks.

261.—THE MONK AND THE BRIDGES.
In this case I give a rough plan of a river with an island and five bridges. On
one side of the river is a monastery, and on the other side is seen a monk in
the foreground. Now, the monk has decided that he will cross every bridge
once, and only once, on his return to the monastery. This is, of course, quite
easy to do, but on the way he thought to himself, "I wonder how many
different routes there are from which I might have selected." Could you
have told him? That is the puzzle. Take your pencil and trace out a route
that will take you once over all the five bridges. Then trace out a second
route, then a third, and see if you can count all the variations. You will find
that the difficulty is twofold: you have to avoid dropping routes on the one
hand and counting the same routes more than once on the other.

COMBINATION AND GROUP PROBLEMS.
"A combination and a form indeed."
Hamlet, iii. 4.
Various puzzles in this class might be termed problems in the "geometry of
situation," but their solution really depends on the theory of combinations
which, in its turn, is derived directly from the theory of permutations. It has
seemed convenient to include here certain group puzzles and enumerations
that might, perhaps, with equal reason have been placed elsewhere; but
readers are again asked not to be too critical about the classification, which
is very difficult and arbitrary. As I have included my problem of "The
Round Table" (No. 273), perhaps a few remarks on another well-known
problem of the same class, known by the French as La Problême des
Ménages, may be interesting. If n married ladies are seated at a round table
in any determined order, in how many different ways may their n husbands
be placed so that every man is between two ladies but never next to his own
wife?
This difficult problem was first solved by Laisant, and the method shown in
the following table is due to Moreau:—

4 0 2
5 3 13
6 13 80
7 83 579
8 592 4738
9 4821 43387
10 43979 439792

The first column shows the number of married couples. The numbers in the
second column are obtained in this way: 5 × 3 + 0 - 2 = 13; 6 × 13 +
3 + 2 = 83; 7 × 83 + 13 - 2 = 592; 8 × 592 + 83 + 2 = 4821; and so on. Find
all the numbers, except 2, in the table, and the method will be evident. It

will be noted that the 2 is subtracted when the first number (the number of
couples) is odd, and added when that number is even. The numbers in the
third column are obtained thus: 13 - 0 = 13; 83 - 3 = 80; 592 - 13 = 579;
4821 - 83 = 4738; and so on. The numbers in this last column give the
required solutions. Thus, four husbands may be seated in two ways, five
husbands may be placed in thirteen ways, and six husbands in eighty ways.
The following method, by Lucas, will show the remarkable way in which
chessboard analysis may be applied to the solution of a circular problem of
this kind. Divide a square into thirty-six cells, six by six, and strike out all
the cells in the long diagonal from the bottom left-hand corner to the top
right-hand corner, also the five cells in the diagonal next above it and the
cell in the bottom right-hand corner. The answer for six couples will be the
same as the number of ways in which you can place six rooks (not using the
cancelled cells) so that no rook shall ever attack another rook. It will be
found that the six rooks may be placed in eighty different ways, which
agrees with the above table.

262.—THOSE FIFTEEN SHEEP.
A certain cyclopædia has the following curious problem, I am told: "Place
fifteen sheep in four pens so that there shall be the same number of sheep in
each pen." No answer whatever is vouchsafed, so I thought I would
investigate the matter. I saw that in dealing with apples or bricks the thing
would appear to be quite impossible, since four times any number must be
an even number, while fifteen is an odd number. I thought, therefore, that
there must be some quality peculiar to the sheep that was not generally
known. So I decided to interview some farmers on the subject. The first one
pointed out that if we put one pen inside another, like the rings of a target,
and placed all sheep in the smallest pen, it would be all right. But I objected
to this, because you admittedly place all the sheep in one pen, not in four
pens. The second man said that if I placed four sheep in each of three pens
and three sheep in the last pen (that is fifteen sheep in all), and one of the
ewes in the last pen had a lamb during the night, there would be the same
number in each pen in the morning. This also failed to satisfy me.

The third farmer said, "I've got four hurdle pens down in one of my fields,
and a small flock of wethers, so if you will just step down with me I will
show you how it is done." The illustration depicts my friend as he is about
to demonstrate the matter to me. His lucid explanation was evidently that
which was in the mind of the writer of the article in the cyclopædia. What
was it? Can you place those fifteen sheep?

263.—KING ARTHUR'S KNIGHTS.
King Arthur sat at the Round Table on three successive evenings with his
knights—Beleobus, Caradoc, Driam, Eric, Floll, and Galahad—but on no
occasion did any person have as his neighbour one who had before sat next
to him. On the first evening they sat in alphabetical order round the table.
But afterwards King Arthur arranged the two next sittings so that he might
have Beleobus as near to him as possible and Galahad as far away from him
as could be managed. How did he seat the knights to the best advantage,
remembering that rule that no knight may have the same neighbour twice?

264.—THE CITY LUNCHEONS.
Twelve men connected with a large firm in the City of London sit down to
luncheon together every day in the same room. The tables are small ones
that only accommodate two persons at the same time. Can you show how
these twelve men may lunch together on eleven days in pairs, so that no two
of them shall ever sit twice together? We will represent the men by the first
twelve letters of the alphabet, and suppose the first day's pairing to be as
follows—
(A B) (C D) (E F) (G H) (I J) (K L).
Then give any pairing you like for the next day, say—
(A C) (B D) (E G) (F H) (I K) (J L),
and so on, until you have completed your eleven lines, with no pair ever
occurring twice. There are a good many different arrangements possible.
Try to find one of them.

265.—A PUZZLE FOR CARD-PLAYERS.
Twelve members of a club arranged to play bridge together on eleven
evenings, but no player was ever to have the same partner more than once,
or the same opponent more than twice. Can you draw up a scheme showing
how they may all sit down at three tables every evening? Call the twelve
players by the first twelve letters of the alphabet and try to group them.

266.—A TENNIS TOURNAMENT.
Four married couples played a "mixed double" tennis tournament, a man
and a lady always playing against a man and a lady. But no person ever

played with or against any other person more than once. Can you show how
they all could have played together in the two courts on three successive
days? This is a little puzzle of a quite practical kind, and it is just perplexing
enough to be interesting.

267.—THE WRONG HATS.
"One of the most perplexing things I have come across lately," said Mr.
Wilson, "is this. Eight men had been dining not wisely but too well at a
certain London restaurant. They were the last to leave, but not one man was
in a condition to identify his own hat. Now, considering that they took their
hats at random, what are the chances that every man took a hat that did not
belong to him?"
"The first thing," said Mr. Waterson, "is to see in how many different ways
the eight hats could be taken."
"That is quite easy," Mr. Stubbs explained. "Multiply together the numbers,
1, 2, 3, 4, 5, 6, 7, and 8. Let me see—half a minute—yes; there are 40,320
different ways."
"Now all you've got to do is to see in how many of these cases no man has
his own hat," said Mr. Waterson.
"Thank you, I'm not taking any," said Mr. Packhurst. "I don't envy the man
who attempts the task of writing out all those forty-thousand-odd cases and
then picking out the ones he wants."
They all agreed that life is not long enough for that sort of amusement; and
as nobody saw any other way of getting at the answer, the matter was
postponed indefinitely. Can you solve the puzzle?

268.—THE PEAL OF BELLS.

A correspondent, who is apparently much interested in campanology, asks
me how he is to construct what he calls a "true and correct" peal for four
bells. He says that every possible permutation of the four bells must be rung
once, and once only. He adds that no bell must move more than one place at
a time, that no bell must make more than two successive strokes in either
the first or the last place, and that the last change must be able to pass into
the first. These fantastic conditions will be found to be observed in the little
peal for three bells, as follows:—

12 3
21 3
23 1
32 1
31 2
13 2

How are we to give him a correct solution for his four bells?

269.—THREE MEN IN A BOAT.
A certain generous London manufacturer gives his workmen every year a
week's holiday at the seaside at his own expense. One year fifteen of his
men paid a visit to Herne Bay. On the morning of their departure from
London they were addressed by their employer, who expressed the hope
that they would have a very pleasant time.
"I have been given to understand," he added, "that some of you fellows are
very fond of rowing, so I propose on this occasion to provide you with this
recreation, and at the same time give you an amusing little puzzle to solve.
During the seven days that you are at Herne Bay every one of you will go
out every day at the same time for a row, but there must always be three
men in a boat and no more. No two men may ever go out in a boat together
more than once, and no man is allowed to go out twice in the same boat. If
you can manage to do this, and use as few different boats as possible, you
may charge the firm with the expense."

One of the men tells me that the experience he has gained in such matters
soon enabled him to work out the answer to the entire satisfaction of
themselves and their employer. But the amusing part of the thing is that
they never really solved the little mystery. I find their method to have been
quite incorrect, and I think it will amuse my readers to discover how the
men should have been placed in the boats. As their names happen to have
been Andrews, Baker, Carter, Danby, Edwards, Frith, Gay, Hart, Isaacs,
Jackson, Kent, Lang, Mason, Napper, and Onslow, we can call them by
their initials and write out the five groups for each of the seven days in the
following simple way:

1 2 3 4 5
First Day: (ABC) (DEF) (GHI) (JKL) (MNO).

The men within each pair of brackets are here seen to be in the same boat,
and therefore A can never go out with B or with C again, and C can never
go out again with B. The same applies to the other four boats. The figures
show the number on the boat, so that A, B, or C, for example, can never go
out in boat No. 1 again.

270.—THE GLASS BALLS.
A number of clever marksmen were staying at a country house, and the
host, to provide a little amusement, suspended strings of glass balls, as
shown in the illustration, to be fired at. After they had all put their skill to a
sufficient test, somebody asked the following question: "What is the total
number of different ways in which these sixteen balls may be broken, if we
must always break the lowest ball that remains on any string?" Thus, one
way would be to break all the four balls on each string in succession, taking
the strings from left to right. Another would be to break all the fourth balls
on the four strings first, then break the three remaining on the first string,
then take the balls on the three other strings alternately from right to left,
and so on. There is such a vast number of different ways (since every little
variation of order makes a different way) that one is apt to be at first
impressed by the great difficulty of the problem. Yet it is really quite simple

when once you have hit on the proper method of attacking it. How many
different ways are there?

271.—FIFTEEN LETTER PUZZLE.

ALE FOE HOD BGN
CAB HEN JOG KFM
HAG GEM MOB BFH
FAN KIN JEK DFL
JAM HIM GCL LJH
AID JIB FCJ NJD
OAK FIG HCK MLN

BED OIL MCD BLK
ICE CON DGK

The above is the solution of a puzzle I gave in Tit-bits in the summer of
1896. It was required to take the letters, A, B, C, D, E, F, G, H, I, J, K, L,
M, N, and O, and with them form thirty-five groups of three letters so that
the combinations should include the greatest number possible of common
English words. No two letters may appear together in a group more than
once. Thus, A and L having been together in ALE, must never be found
together again; nor may A appear again in a group with E, nor L with E.
These conditions will be found complied with in the above solution, and the
number of words formed is twenty-one. Many persons have since tried hard
to beat this number, but so far have not succeeded.
More than thirty-five combinations of the fifteen letters cannot be formed
within the conditions. Theoretically, there cannot possibly be more than
twenty-three words formed, because only this number of combinations is
possible with a vowel or vowels in each. And as no English word can be
formed from three of the given vowels (A, E, I, and O), we must reduce the
number of possible words to twenty-two. This is correct theoretically, but
practically that twenty-second word cannot be got in. If JEK, shown above,
were a word it would be all right; but it is not, and no amount of juggling
with the other letters has resulted in a better answer than the one shown. I
should, say that proper nouns and abbreviations, such as Joe, Jim, Alf, Hal,
Flo, Ike, etc., are disallowed.
Now, the present puzzle is a variation of the above. It is simply this: Instead
of using the fifteen letters given, the reader is allowed to select any fifteen
different letters of the alphabet that he may prefer. Then construct thirty-
five groups in accordance with the conditions, and show as many good
English words as possible.

272.—THE NINE SCHOOLBOYS.

This is a new and interesting companion puzzle to the "Fifteen Schoolgirls"
(see solution of No. 269), and even in the simplest possible form in which I
present it there are unquestionable difficulties. Nine schoolboys walk out in
triplets on the six week days so that no boy ever walks side by side with any
other boy more than once. How would you arrange them?
If we represent them by the first nine letters of the alphabet, they might be
grouped on the first day as follows:—

ABC
DE F
GH I

Then A can never walk again side by side with B, or B with C, or D with E,
and so on. But A can, of course, walk side by side with C. It is here not a
question of being together in the same triplet, but of walking side by side in
a triplet. Under these conditions they can walk out on six days; under the
"Schoolgirls" conditions they can only walk on four days.

273.—THE ROUND TABLE.
Seat the same n persons at a round table on

(n - 1)(n - 2)
2

occasions so that no person shall ever have the same two neighbours twice.
This is, of course, equivalent to saying that every person must sit once, and
once only, between every possible pair.

274.—THE MOUSE-TRAP PUZZLE.

This is a modern version, with a difference, of an old puzzle of the same
name. Number twenty-one cards, 1, 2, 3, etc., up to 21, and place them in a
circle in the particular order shown in the illustration. These cards represent
mice. You start from any card, calling that card "one," and count, "one, two,
three," etc., in a clockwise direction, and when your count agrees with the
number on the card, you have made a "catch," and you remove the card.
Then start at the next card, calling that "one," and try again to make another
"catch." And so on. Supposing you start at 18, calling that card "one," your
first "catch" will be 19. Remove 19 and your next "catch" is 10. Remove 10
and your next "catch" is 1. Remove the 1, and if you count up to 21 (you
must never go beyond), you cannot make another "catch." Now, the ideal is
to "catch" all the twenty-one mice, but this is not here possible, and if it
were it would merely require twenty-one different trials, at the most, to
succeed. But the reader may make any two cards change places before he
begins. Thus, you can change the 6 with the 2, or the 7 with the 11, or any
other pair. This can be done in several ways so as to enable you to "catch"
all the twenty-one mice, if you then start at the right place. You may never
pass over a "catch"; you must always remove the card and start afresh.

275.—THE SIXTEEN SHEEP.

Here is a new puzzle with matches and counters or coins. In the illustration
the matches represent hurdles and the counters sheep. The sixteen hurdles
on the outside, and the sheep, must be regarded as immovable; the puzzle
has to do entirely with the nine hurdles on the inside. It will be seen that at
present these nine hurdles enclose four groups of 8, 3, 3, and 2 sheep. The
farmer requires to readjust some of the hurdles so as to enclose 6, 6, and 4
sheep. Can you do it by only replacing two hurdles? When you have
succeeded, then try to do it by replacing three hurdles; then four, five, six,
and seven in succession. Of course, the hurdles must be legitimately laid on
the dotted lines, and no such tricks are allowed as leaving unconnected ends
of hurdles, or two hurdles placed side by side, or merely making hurdles
change places. In fact, the conditions are so simple that any farm labourer
will understand it directly.

276.—THE EIGHT VILLAS.
In one of the outlying suburbs of London a man had a square plot of ground
on which he decided to build eight villas, as shown in the illustration, with a
common recreation ground in the middle. After the houses were completed,
and all or some of them let, he discovered that the number of occupants in
the three houses forming a side of the square was in every case nine. He did
not state how the occupants were distributed, but I have shown by the
numbers on the sides of the houses one way in which it might have
happened. The puzzle is to discover the total number of ways in which all
or any of the houses might be occupied, so that there should be nine persons
on each side. In order that there may be no misunderstanding, I will explain
that although B is what we call a reflection of A, these would count as two
different arrangements, while C, if it is turned round, will give four
arrangements; and if turned round in front of a mirror, four other
arrangements. All eight must be counted.

277.—COUNTER CROSSES.
All that we need for this puzzle is nine counters, numbered 1, 2, 3, 4, 5, 6,
7, 8, and 9. It will be seen that in the illustration A these are arranged so as
to form a Greek cross, while in the case of B they form a Latin cross. In
both cases the reader will find that the sum of the numbers in the upright of
the cross is the same as the sum of the numbers in the horizontal arm. It is
quite easy to hit on such an arrangement by trial, but the problem is to
discover in exactly how many different ways it may be done in each case.
Remember that reversals and reflections do not count as different. That is to

say, if you turn this page round you get four arrangements of the Greek
cross, and if you turn it round again in front of a mirror you will get four
more. But these eight are all regarded as one and the same. Now, how many
different ways are there in each case?

278.—A DORMITORY PUZZLE.
In a certain convent there were eight large dormitories on one floor,
approached by a spiral staircase in the centre, as shown in our plan. On an
inspection one Monday by the abbess it was found that the south aspect was
so much preferred that six times as many nuns slept on the south side as on
each of the other three sides. She objected to this overcrowding, and
ordered that it should be reduced. On Tuesday she found that five times as
many slept on the south side as on each of the other sides. Again she
complained. On Wednesday she found four times as many on the south side,
on Thursday three times as many, and on Friday twice as many. Urging the
nuns to further efforts, she was pleased to find on Saturday that an equal
number slept on each of the four sides of the house. What is the smallest
number of nuns there could have been, and how might they have arranged
themselves on each of the six nights? No room may ever be unoccupied.

279.—THE BARRELS OF BALSAM.

A merchant of Bagdad had ten barrels of precious balsam for sale. They
were numbered, and were arranged in two rows, one on top of the other, as
shown in the picture. The smaller the number on the barrel, the greater was
its value. So that the best quality was numbered "1" and the worst numbered
"10," and all the other numbers of graduating values. Now, the rule of
Ahmed Assan, the merchant, was that he never put a barrel either beneath or
to the right of one of less value. The arrangement shown is, of course, the
simplest way of complying with this condition. But there are many other
ways—such, for example, as this:—

125 7 8
346 910

Here, again, no barrel has a smaller number than itself on its right or
beneath it. The puzzle is to discover in how many different ways the
merchant of Bagdad might have arranged his barrels in the two rows
without breaking his rule. Can you count the number of ways?

280.—BUILDING THE TETRAHEDRON.
I possess a tetrahedron, or triangular pyramid, formed of six sticks glued
together, as shown in the illustration. Can you count correctly the number of
different ways in which these six sticks might have been stuck together so
as to form the pyramid?
Some friends worked at it together one evening, each person providing
himself with six lucifer matches to aid his thoughts; but it was found that no
two results were the same. You see, if we remove one of the sticks and turn
it round the other way, that will be a different pyramid. If we make two of
the sticks change places the result will again be different. But remember
that every pyramid may be made to stand on either of its four sides without
being a different one. How many ways are there altogether?

281.—PAINTING A PYRAMID.

This puzzle concerns the painting of the four sides of a tetrahedron, or
triangular pyramid. If you cut out a piece of cardboard of the triangular
shape shown in Fig. 1, and then cut half through along the dotted lines, it
will fold up and form a perfect triangular pyramid. And I would first remind
my readers that the primary colours of the solar spectrum are seven—violet,
indigo, blue, green, yellow, orange, and red. When I was a child I was
taught to remember these by the ungainly word formed by the initials of the
colours, "Vibgyor."

In how many different ways may the triangular pyramid be coloured, using
in every case one, two, three, or four colours of the solar spectrum? Of
course a side can only receive a single colour, and no side can be left
uncoloured. But there is one point that I must make quite clear. The four
sides are not to be regarded as individually distinct. That is to say, if you
paint your pyramid as shown in Fig. 2 (where the bottom side is green and
the other side that is out of view is yellow), and then paint another in the
order shown in Fig. 3, these are really both the same and count as one way.
For if you tilt over No. 2 to the right it will so fall as to represent No. 3. The
avoidance of repetitions of this kind is the real puzzle of the thing. If a
coloured pyramid cannot be placed so that it exactly resembles in its
colours and their relative order another pyramid, then they are different.
Remember that one way would be to colour all the four sides red, another to
colour two sides green, and the remaining sides yellow and blue; and so on.

282.—THE ANTIQUARY'S CHAIN.
An antiquary possessed a number of curious old links, which he took to a
blacksmith, and told him to join together to form one straight piece of
chain, with the sole condition that the two circular links were not to be
together. The following illustration shows the appearance of the chain and
the form of each link. Now, supposing the owner should separate the links
again, and then take them to another smith and repeat his former
instructions exactly, what are the chances against the links being put
together exactly as they were by the first man? Remember that every
successive link can be joined on to another in one of two ways, just as you
can put a ring on your finger in two ways, or link your forefingers and
thumbs in two ways.

283.—THE FIFTEEN DOMINOES.
In this case we do not use the complete set of twenty-eight dominoes to be
found in the ordinary box. We dispense with all those dominoes that have a
five or a six on them and limit ourselves to the fifteen that remain, where
the double-four is the highest.
In how many different ways may the fifteen dominoes be arranged in a
straight line in accordance with the simple rule of the game that a number
must always be placed against a similar number—that is, a four against a
four, a blank against a blank, and so on? Left to right and right to left of the
same arrangement are to be counted as two different ways.

284.—THE CROSS TARGET.

In the illustration we have a somewhat curious target designed by an
eccentric sharpshooter. His idea was that in order to score you must hit four
circles in as many shots so that those four shots shall form a square. It will
be seen by the results recorded on the target that two attempts have been
successful. The first man hit the four circles at the top of the cross, and thus
formed his square. The second man intended to hit the four in the bottom
arm, but his second shot, on the left, went too high. This compelled him to
complete his four in a different way than he intended. It will thus be seen
that though it is immaterial which circle you hit at the first shot, the second
shot may commit you to a definite procedure if you are to get your square.
Now, the puzzle is to say in just how many different ways it is possible to
form a square on the target with four shots.

285.—THE FOUR POSTAGE STAMPS.

"It is as easy as counting," is an expression one sometimes hears. But mere
counting may be puzzling at times. Take the following simple example.
Suppose you have just bought twelve postage stamps, in this form—three
by four—and a friend asks you to oblige him with four stamps, all joined
together—no stamp hanging on by a mere corner. In how many different
ways is it possible for you to tear off those four stamps? You see, you can
give him 1, 2, 3, 4, or 2, 3, 6, 7, or 1, 2, 3, 6, or 1, 2, 3, 7, or 2, 3, 4, 8, and
so on. Can you count the number of different ways in which those four
stamps might be delivered? There are not many more than fifty ways, so it
is not a big count. Can you get the exact number?

286.—PAINTING THE DIE.
In how many different ways may the numbers on a single die be marked,
with the only condition that the 1 and 6, the 2 and 5, and the 3 and 4 must
be on opposite sides? It is a simple enough question, and yet it will puzzle a
good many people.

287.—AN ACROSTIC PUZZLE.
In the making or solving of double acrostics, has it ever occurred to you to
consider the variety and limitation of the pair of initial and final letters
available for cross words? You may have to find a word beginning with A
and ending with B, or A and C, or A and D, and so on. Some combinations
are obviously impossible—such, for example, as those with Q at the end.
But let us assume that a good English word can be found for every case.
Then how many possible pairs of letters are available?

CHESSBOARD PROBLEMS.
"You and I will goe to the chesse."
GREENE'S Groatsworth of Wit.
During a heavy gale a chimney-pot was hurled through the air, and crashed
upon the pavement just in front of a pedestrian. He quite calmly said, "I
have no use for it: I do not smoke." Some readers, when they happen to see
a puzzle represented on a chessboard with chess pieces, are apt to make the
equally inconsequent remark, "I have no use for it: I do not play chess."
This is largely a result of the common, but erroneous, notion that the
ordinary chess puzzle with which we are familiar in the press (dignified, for
some reason, with the name "problem") has a vital connection with the
game of chess itself. But there is no condition in the game that you shall
checkmate your opponent in two moves, in three moves, or in four moves,
while the majority of the positions given in these puzzles are such that one
player would have so great a superiority in pieces that the other would have
resigned before the situations were reached. And the solving of them helps
you but little, and that quite indirectly, in playing the game, it being well
known that, as a rule, the best "chess problemists" are indifferent players,
and vice versa. Occasionally a man will be found strong on both subjects,
but he is the exception to the rule.
Yet the simple chequered board and the characteristic moves of the pieces
lend themselves in a very remarkable manner to the devising of the most
entertaining puzzles. There is room for such infinite variety that the true
puzzle lover cannot afford to neglect them. It was with a view to securing
the interest of readers who are frightened off by the mere presentation of a
chessboard that so many puzzles of this class were originally published by
me in various fanciful dresses. Some of these posers I still retain in their
disguised form; others I have translated into terms of the chessboard. In the
majority of cases the reader will not need any knowledge whatever of chess,
but I have thought it best to assume throughout that he is acquainted with
the terminology, the moves, and the notation of the game.

I first deal with a few questions affecting the chessboard itself; then with
certain statical puzzles relating to the Rook, the Bishop, the Queen, and the
Knight in turn; then dynamical puzzles with the pieces in the same order;
and, finally, with some miscellaneous puzzles on the chessboard. It is hoped
that the formulæ and tables given at the end of the statical puzzles will be of
interest, as they are, for the most part, published for the first time.

THE CHESSBOARD.
"Good company's a chessboard."
BYRON'S Don Juan, xiii. 89.
A chessboard is essentially a square plane divided into sixty-four smaller
squares by straight lines at right angles. Originally it was not chequered
(that is, made with its rows and columns alternately black and white, or of
any other two colours), and this improvement was introduced merely to
help the eye in actual play. The utility of the chequers is unquestionable.
For example, it facilitates the operation of the bishops, enabling us to see at
the merest glance that our king or pawns on black squares are not open to
attack from an opponent's bishop running on the white diagonals. Yet the
chequering of the board is not essential to the game of chess. Also, when we
are propounding puzzles on the chessboard, it is often well to remember
that additional interest may result from "generalizing" for boards containing
any number of squares, or from limiting ourselves to some particular
chequered arrangement, not necessarily a square. We will give a few
puzzles dealing with chequered boards in this general way.

288.—CHEQUERED BOARD DIVISIONS.
I recently asked myself the question: In how many different ways may a
chessboard be divided into two parts of the same size and shape by cuts
along the lines dividing the squares? The problem soon proved to be both
fascinating and bristling with difficulties. I present it in a simplified form,
taking a board of smaller dimensions.

It is obvious that a board of four squares can only be so divided in one way
—by a straight cut down the centre—because we shall not count reversals
and reflections as different. In the case of a board of sixteen squares—four
by four—there are just six different ways. I have given all these in the
diagram, and the reader will not find any others. Now, take the larger board
of thirty-six squares, and try to discover in how many ways it may be cut
into two parts of the same size and shape.

289.—LIONS AND CROWNS.
The young lady in the illustration is confronted with a little cutting-out
difficulty in which the reader may be glad to assist her. She wishes, for
some reason that she has not communicated to me, to cut that square piece
of valuable material into four parts, all of exactly the same size and shape,
but it is important that every piece shall contain a lion and a crown. As she
insists that the cuts can only be made along the lines dividing the squares,
she is considerably perplexed to find out how it is to be done. Can you
show her the way? There is only one possible method of cutting the stuff.

290.—BOARDS WITH AN ODD NUMBER OF SQUARES.

We will here consider the question of those boards that contain an odd
number of squares. We will suppose that the central square is first cut out,
so as to leave an even number of squares for division. Now, it is obvious
that a square three by three can only be divided in one way, as shown in
Fig. 1. It will be seen that the pieces A and B are of the same size and shape,
and that any other way of cutting would only produce the same shaped
pieces, so remember that these variations are not counted as different ways.
The puzzle I propose is to cut the board five by five (Fig. 2) into two pieces
of the same size and shape in as many different ways as possible. I have
shown in the illustration one way of doing it. How many different ways are
there altogether? A piece which when turned over resembles another piece
is not considered to be of a different shape.

291.—THE GRAND LAMA'S PROBLEM.
Once upon a time there was a Grand Lama who had a chessboard made of
pure gold, magnificently engraved, and, of course, of great value. Every

year a tournament was held at Lhassa among the priests, and whenever any
one beat the Grand Lama it was considered a great honour, and his name
was inscribed on the back of the board, and a costly jewel set in the
particular square on which the checkmate had been given. After this
sovereign pontiff had been defeated on four occasions he died—possibly of
chagrin.

Now the new Grand Lama was an inferior chess-player, and preferred other
forms of innocent amusement, such as cutting off people's heads. So he
discouraged chess as a degrading game, that did not improve either the
mind or the morals, and abolished the tournament summarily. Then he sent
for the four priests who had had the effrontery to play better than a Grand
Lama, and addressed them as follows: "Miserable and heathenish men,
calling yourselves priests! Know ye not that to lay claim to a capacity to do
anything better than my predecessor is a capital offence? Take that
chessboard and, before day dawns upon the torture chamber, cut it into four
equal parts of the same shape, each containing sixteen perfect squares, with
one of the gems in each part! If in this you fail, then shall other sports be
devised for your special delectation. Go!" The four priests succeeded in

their apparently hopeless task. Can you show how the board may be divided
into four equal parts, each of exactly the same shape, by cuts along the lines
dividing the squares, each part to contain one of the gems?

292.—THE ABBOT'S WINDOW.
Once upon a time the Lord Abbot of St. Edmondsbury, in consequence of
"devotions too strong for his head," fell sick and was unable to leave his
bed. As he lay awake, tossing his head restlessly from side to side, the
attentive monks noticed that something was disturbing his mind; but
nobody dared ask what it might be, for the abbot was of a stern disposition,
and never would brook inquisitiveness. Suddenly he called for Father John,
and that venerable monk was soon at the bedside.
"Father John," said the Abbot, "dost thou know that I came into this wicked
world on a Christmas Even?"
The monk nodded assent.
"And have I not often told thee that, having been born on Christmas Even, I
have no love for the things that are odd? Look there!"

The Abbot pointed to the large dormitory window, of which I give a sketch.
The monk looked, and was perplexed.
"Dost thou not see that the sixty-four lights add up an even number
vertically and horizontally, but that all the diagonal lines, except fourteen
are of a number that is odd? Why is this?"
"Of a truth, my Lord Abbot, it is of the very nature of things, and cannot be
changed."
"Nay, but it shall be changed. I command thee that certain of the lights be
closed this day, so that every line shall have an even number of lights. See
thou that this be done without delay, lest the cellars be locked up for a
month and other grievous troubles befall thee."
Father John was at his wits' end, but after consultation with one who was
learned in strange mysteries, a way was found to satisfy the whim of the
Lord Abbot. Which lights were blocked up, so that those which remained
added up an even number in every line horizontally, vertically, and
diagonally, while the least possible obstruction of light was caused?

293.—THE CHINESE CHESSBOARD.
Into how large a number of different pieces may the chessboard be cut (by
cuts along the lines only), no two pieces being exactly alike? Remember
that the arrangement of black and white constitutes a difference. Thus, a
single black square will be different from a single white square, a row of
three containing two white squares will differ from a row of three
containing two black, and so on. If two pieces cannot be placed on the table
so as to be exactly alike, they count as different. And as the back of the
board is plain, the pieces cannot be turned over.

294.—THE CHESSBOARD SENTENCE.

I once set myself the amusing task of so dissecting an ordinary chessboard
into letters of the alphabet that they would form a complete sentence. It will
be seen from the illustration that the pieces assembled give the sentence,
"CUT THY LIFE," with the stops between. The ideal sentence would, of
course, have only one full stop, but that I did not succeed in obtaining.
The sentence is an appeal to the transgressor to cut himself adrift from the
evil life he is living. Can you fit these pieces together to form a perfect
chessboard?

STATICAL CHESS PUZZLES.
"They also serve who only stand and wait."
MILTON.

295.—THE EIGHT ROOKS.

It will be seen in the first diagram that every square on the board is either
occupied or attacked by a rook, and that every rook is "guarded" (if they
were alternately black and white rooks we should say "attacked") by
another rook. Placing the eight rooks on any row or file obviously will have
the same effect. In diagram 2 every square is again either occupied or
attacked, but in this case every rook is unguarded. Now, in how many
different ways can you so place the eight rooks on the board that every

square shall be occupied or attacked and no rook ever guarded by another? I
do not wish to go into the question of reversals and reflections on this
occasion, so that placing the rooks on the other diagonal will count as
different, and similarly with other repetitions obtained by turning the board
round.

296.—THE FOUR LIONS.
The puzzle is to find in how many different ways the four lions may be
placed so that there shall never be more than one lion in any row or column.
Mere reversals and reflections will not count as different. Thus, regarding
the example given, if we place the lions in the other diagonal, it will be
considered the same arrangement. For if you hold the second arrangement
in front of a mirror or give it a quarter turn, you merely get the first
arrangement. It is a simple little puzzle, but requires a certain amount of
careful consideration.

297.—BISHOPS—UNGUARDED.
Place as few bishops as possible on an ordinary chessboard so that every
square of the board shall be either occupied or attacked. It will be seen that
the rook has more scope than the bishop: for wherever you place the former,
it will always attack fourteen other squares; whereas the latter will attack
seven, nine, eleven, or thirteen squares, according to the position of the
diagonal on which it is placed. And it is well here to state that when we
speak of "diagonals" in connection with the chessboard, we do not limit
ourselves to the two long diagonals from corner to corner, but include all
the shorter lines that are parallel to these. To prevent misunderstanding on
future occasions, it will be well for the reader to note carefully this fact.

298.—BISHOPS—GUARDED.

Now, how many bishops are necessary in order that every square shall be
either occupied or attacked, and every bishop guarded by another bishop?
And how may they be placed?

299.—BISHOPS IN CONVOCATION.

The greatest number of bishops that can be placed at the same time on the
chessboard, without any bishop attacking another, is fourteen. I show, in
diagram, the simplest way of doing this. In fact, on a square chequered
board of any number of squares the greatest number of bishops that can be
placed without attack is always two less than twice the number of squares
on the side. It is an interesting puzzle to discover in just how many different
ways the fourteen bishops may be so placed without mutual attack. I shall
give an exceedingly simple rule for determining the number of ways for a
square chequered board of any number of squares.

300.—THE EIGHT QUEENS.
The queen is by far the strongest piece on the chessboard. If you place her
on one of the four squares in the centre of the board, she attacks no fewer
than twenty-seven other squares; and if you try to hide her in a corner, she
still attacks twenty-one squares. Eight queens may be placed on the board
so that no queen attacks another, and it is an old puzzle (first proposed by
Nauck in 1850, and it has quite a little literature of its own) to discover in
just how many different ways this may be done. I show one way in the
diagram, and there are in all twelve of these fundamentally different ways.
These twelve produce ninety-two ways if we regard reversals and
reflections as different. The diagram is in a way a symmetrical arrangement.
If you turn the page upside down, it will reproduce itself exactly; but if you
look at it with one of the other sides at the bottom, you get another way that
is not identical. Then if you reflect these two ways in a mirror you get two
more ways. Now, all the other eleven solutions are non-symmetrical, and
therefore each of them may be presented in eight ways by these reversals
and reflections. It will thus be seen why the twelve fundamentally different
solutions produce only ninety-two arrangements, as I have said, and not
ninety-six, as would happen if all twelve were non-symmetrical. It is well to
have a clear understanding on the matter of reversals and reflections when
dealing with puzzles on the chessboard.

Can the reader place the eight queens on the board so that no queen shall
attack another and so that no three queens shall be in a straight line in any
oblique direction? Another glance at the diagram will show that this
arrangement will not answer the conditions, for in the two directions
indicated by the dotted lines there are three queens in a straight line. There
is only one of the twelve fundamental ways that will solve the puzzle. Can
you find it?

301.—THE EIGHT STARS.

The puzzle in this case is to place eight stars in the diagram so that no star
shall be in line with another star horizontally, vertically, or diagonally. One
star is already placed, and that must not be moved, so there are only seven
for the reader now to place. But you must not place a star on any one of the
shaded squares. There is only one way of solving this little puzzle.

302.—A PROBLEM IN MOSAICS.
The art of producing pictures or designs by means of joining together pieces
of hard substances, either naturally or artificially coloured, is of very great
antiquity. It was certainly known in the time of the Pharaohs, and we find a
reference in the Book of Esther to "a pavement of red, and blue, and white,
and black marble." Some of this ancient work that has come down to us,
especially some of the Roman mosaics, would seem to show clearly, even
where design is not at first evident, that much thought was bestowed upon
apparently disorderly arrangements. Where, for example, the work has been
produced with a very limited number of colours, there are evidences of

great ingenuity in preventing the same tints coming in close proximity.
Lady readers who are familiar with the construction of patchwork quilts
will know how desirable it is sometimes, when they are limited in the
choice of material, to prevent pieces of the same stuff coming too near
together. Now, this puzzle will apply equally to patchwork quilts or
tesselated pavements.
It will be seen from the diagram how a square piece of flooring may be
paved with sixty-two square tiles of the eight colours violet, red, yellow,
green, orange, purple, white, and blue (indicated by the initial letters), so
that no tile is in line with a similarly coloured tile, vertically, horizontally,
or diagonally. Sixty-four such tiles could not possibly be placed under these
conditions, but the two shaded squares happen to be occupied by iron
ventilators.

The puzzle is this. These two ventilators have to be removed to the
positions indicated by the darkly bordered tiles, and two tiles placed in
those bottom corner squares. Can you readjust the thirty-two tiles so that no
two of the same colour shall still be in line?

303.—UNDER THE VEIL.

If the reader will examine the above diagram, he will see that I have so
placed eight V's, eight E's, eight I's, and eight L's in the diagram that no
letter is in line with a similar one horizontally, vertically, or diagonally.
Thus, no V is in line with another V, no E with another E, and so on. There
are a great many different ways of arranging the letters under this condition.
The puzzle is to find an arrangement that produces the greatest possible
number of four-letter words, reading upwards and downwards, backwards
and forwards, or diagonally. All repetitions count as different words, and the
five variations that may be used are: VEIL, VILE, LEVI, LIVE, and EVIL.
This will be made perfectly clear when I say that the above arrangement
scores eight, because the top and bottom row both give VEIL; the second
and seventh columns both give VEIL; and the two diagonals, starting from
the L in the 5th row and E in the 8th row, both give LIVE and EVIL. There
are therefore eight different readings of the words in all.

This difficult word puzzle is given as an example of the use of chessboard
analysis in solving such things. Only a person who is familiar with the
"Eight Queens" problem could hope to solve it.

304.—BACHET'S SQUARE.
One of the oldest card puzzles is by Claude Caspar Bachet de Méziriac, first
published, I believe, in the 1624 edition of his work. Rearrange the sixteen
court cards (including the aces) in a square so that in no row of four cards,
horizontal, vertical, or diagonal, shall be found two cards of the same suit or
the same value. This in itself is easy enough, but a point of the puzzle is to
find in how many different ways this may be done. The eminent French
mathematician A. Labosne, in his modern edition of Bachet, gives the
answer incorrectly. And yet the puzzle is really quite easy. Any arrangement
produces seven more by turning the square round and reflecting it in a
mirror. These are counted as different by Bachet.
Note "row of four cards," so that the only diagonals we have here to
consider are the two long ones.

305.—THE THIRTY-SIX LETTER BLOCKS.

The illustration represents a box containing thirty-six letter-blocks. The
puzzle is to rearrange these blocks so that no A shall be in a line vertically,
horizontally, or diagonally with another A, no B with another B, no C with
another C, and so on. You will find it impossible to get all the letters into
the box under these conditions, but the point is to place as many as possible.
Of course no letters other than those shown may be used.

306.—THE CROWDED CHESSBOARD.

The puzzle is to rearrange the fifty-one pieces on the chessboard so that no
queen shall attack another queen, no rook attack another rook, no bishop
attack another bishop, and no knight attack another knight. No notice is to
be taken of the intervention of pieces of another type from that under
consideration—that is, two queens will be considered to attack one another
although there may be, say, a rook, a bishop, and a knight between them.
And so with the rooks and bishops. It is not difficult to dispose of each type
of piece separately; the difficulty comes in when you have to find room for
all the arrangements on the board simultaneously.

307.—THE COLOURED COUNTERS.

The diagram represents twenty-five coloured counters, Red, Blue, Yellow,
Orange, and Green (indicated by their initials), and there are five of each
colour, numbered 1, 2, 3, 4, and 5. The problem is so to place them in a
square that neither colour nor number shall be found repeated in any one of
the five rows, five columns, and two diagonals. Can you so rearrange them?

308.—THE GENTLE ART OF STAMP-LICKING.
The Insurance Act is a most prolific source of entertaining puzzles,
particularly entertaining if you happen to be among the exempt. One's
initiation into the gentle art of stamp-licking suggests the following little
poser: If you have a card divided into sixteen spaces (4 × 4), and are
provided with plenty of stamps of the values 1d., 2d., 3d., 4d., and 5d., what
is the greatest value that you can stick on the card if the Chancellor of the
Exchequer forbids you to place any stamp in a straight line (that is,
horizontally, vertically, or diagonally) with another stamp of similar value?

Of course, only one stamp can be affixed in a space. The reader will
probably find, when he sees the solution, that, like the stamps themselves,
he is licked He will most likely be twopence short of the maximum. A
friend asked the Post Office how it was to be done; but they sent him to the
Customs and Excise officer, who sent him to the Insurance Commissioners,
who sent him to an approved society, who profanely sent him—but no
matter.

309.—THE FORTY-NINE COUNTERS.

Can you rearrange the above forty-nine counters in a square so that no
letter, and also no number, shall be in line with a similar one, vertically,
horizontally, or diagonally? Here I, of course, mean in the lines parallel
with the diagonals, in the chessboard sense.

310.—THE THREE SHEEP.

A farmer had three sheep and an arrangement of sixteen pens, divided off
by hurdles in the manner indicated in the illustration. In how many different
ways could he place those sheep, each in a separate pen, so that every pen
should be either occupied or in line (horizontally, vertically, or diagonally)
with at least one sheep? I have given one arrangement that fulfils the
conditions. How many others can you find? Mere reversals and reflections
must not be counted as different. The reader may regard the sheep as
queens. The problem is then to place the three queens so that every square
shall be either occupied or attacked by at least one queen—in the maximum
number of different ways.

311.—THE FIVE DOGS PUZZLE.
In 1863, C.F. de Jaenisch first discussed the "Five Queens Puzzle"—to
place five queens on the chessboard so that every square shall be attacked

or occupied—which was propounded by his friend, a "Mr. de R." Jaenisch
showed that if no queen may attack another there are ninety-one different
ways of placing the five queens, reversals and reflections not counting as
different. If the queens may attack one another, I have recorded hundreds of
ways, but it is not practicable to enumerate them exactly.

The illustration is supposed to represent an arrangement of sixty-four
kennels. It will be seen that five kennels each contain a dog, and on further
examination it will be seen that every one of the sixty-four kennels is in a
straight line with at least one dog—either horizontally, vertically, or
diagonally. Take any kennel you like, and you will find that you can draw a
straight line to a dog in one or other of the three ways mentioned. The
puzzle is to replace the five dogs and discover in just how many different
ways they may be placed in five kennels in a straight row, so that every
kennel shall always be in line with at least one dog. Reversals and
reflections are here counted as different.

312.—THE FIVE CRESCENTS OF BYZANTIUM.
When Philip of Macedon, the father of Alexander the Great, found himself
confronted with great difficulties in the siege of Byzantium, he set his men
to undermine the walls. His desires, however, miscarried, for no sooner had
the operations been begun than a crescent moon suddenly appeared in the
heavens and discovered his plans to his adversaries. The Byzantines were
naturally elated, and in order to show their gratitude they erected a statue to
Diana, and the crescent became thenceforward a symbol of the state. In the
temple that contained the statue was a square pavement composed of sixty-
four large and costly tiles. These were all plain, with the exception of five,
which bore the symbol of the crescent. These five were for occult reasons
so placed that every tile should be watched over by (that is, in a straight
line, vertically, horizontally, or diagonally with) at least one of the
crescents. The arrangement adopted by the Byzantine architect was as
follows:—

Now, to cover up one of these five crescents was a capital offence, the death
being something very painful and lingering. But on a certain occasion of

festivity it was necessary to lay down on this pavement a square carpet of
the largest dimensions possible, and I have shown in the illustration by dark
shading the largest dimensions that would be available.
The puzzle is to show how the architect, if he had foreseen this question of
the carpet, might have so arranged his five crescent tiles in accordance with
the required conditions, and yet have allowed for the largest possible square
carpet to be laid down without any one of the five crescent tiles being
covered, or any portion of them.

313.—QUEENS AND BISHOP PUZZLE.
It will be seen that every square of the board is either occupied or attacked.
The puzzle is to substitute a bishop for the rook on the same square, and
then place the four queens on other squares so that every square shall again
be either occupied or attacked.

314.—THE SOUTHERN CROSS.

In the above illustration we have five Planets and eighty-one Fixed Stars,
five of the latter being hidden by the Planets. It will be found that every
Star, with the exception of the ten that have a black spot in their centres, is
in a straight line, vertically, horizontally, or diagonally, with at least one of
the Planets. The puzzle is so to rearrange the Planets that all the Stars shall
be in line with one or more of them.
In rearranging the Planets, each of the five may be moved once in a straight
line, in either of the three directions mentioned. They will, of course,
obscure five other Stars in place of those at present covered.

315.—THE HAT-PEG PUZZLE.

Here is a five-queen puzzle that I gave in a fanciful dress in 1897. As the
queens were there represented as hats on sixty-four pegs, I will keep to the
title, "The Hat-Peg Puzzle." It will be seen that every square is occupied or
attacked. The puzzle is to remove one queen

to a different square so that still every square is occupied or attacked, then
move a second queen under a similar condition, then a third queen, and
finally a fourth queen. After the fourth move every square must be attacked
or occupied, but no queen must then attack another. Of course, the moves
need not be "queen moves;" you can move a queen to any part of the board.

316.—THE AMAZONS.

This puzzle is based on one by Captain Turton. Remove three of the queens
to other squares so that there shall be eleven squares on the board that are
not attacked. The removal of the three queens need not be by "queen
moves." You may take them up and place them anywhere. There is only one
solution.

317.—A PUZZLE WITH PAWNS.
Place two pawns in the middle of the chessboard, one at Q 4 and the other
at K 5. Now, place the remaining fourteen pawns (sixteen in all) so that no
three shall be in a straight line in any possible direction.
Note that I purposely do not say queens, because by the words "any
possible direction" I go beyond attacks on diagonals. The pawns must be
regarded as mere points in space—at the centres of the squares. See dotted
lines in the case of No. 300, "The Eight Queens."

318.—LION-HUNTING.

My friend Captain Potham Hall, the renowned hunter of big game, says
there is nothing more exhilarating than a brush with a herd—a pack—a
team—a flock—a swarm (it has taken me a full quarter of an hour to recall
the right word, but I have it at last)—a pride of lions. Why a number of
lions are called a "pride," a number of whales a "school," and a number of
foxes a "skulk" are mysteries of philology into which I will not enter.
Well, the captain says that if a spirited lion crosses your path in the desert it
becomes lively, for the lion has generally been looking for the man just as
much as the man has sought the king of the forest. And yet when they meet
they always quarrel and fight it out. A little contemplation of this
unfortunate and long-standing feud between two estimable families has led
me to figure out a few calculations as to the probability of the man and the
lion crossing one another's path in the jungle. In all these cases one has to
start on certain more or less arbitrary assumptions. That is why in the above
illustration I have thought it necessary to represent the paths in the desert

with such rigid regularity. Though the captain assures me that the tracks of
the lions usually run much in this way, I have doubts.
The puzzle is simply to find out in how many different ways the man and
the lion may be placed on two different spots that are not on the same path.
By "paths" it must be understood that I only refer to the ruled lines. Thus,
with the exception of the four corner spots, each combatant is always on
two paths and no more. It will be seen that there is a lot of scope for
evading one another in the desert, which is just what one has always
understood.

319.—THE KNIGHT-GUARDS.

The knight is the irresponsible low comedian of the chessboard. "He is a
very uncertain, sneaking, and demoralizing rascal," says an American
writer. "He can only move two squares, but makes up in the quality of his
locomotion for its quantity, for he can spring one square sideways and one

forward simultaneously, like a cat; can stand on one leg in the middle of the
board and jump to any one of eight squares he chooses; can get on one side
of a fence and blackguard three or four men on the other; has an
objectionable way of inserting himself in safe places where he can scare the
king and compel him to move, and then gobble a queen. For pure
cussedness the knight has no equal, and when you chase him out of one
hole he skips into another." Attempts have been made over and over again
to obtain a short, simple, and exact definition of the move of the knight—
without success. It really consists in moving one square like a rook, and
then another square like a bishop—the two operations being done in one
leap, so that it does not matter whether the first square passed over is
occupied by another piece or not. It is, in fact, the only leaping move in
chess. But difficult as it is to define, a child can learn it by inspection in a
few minutes.
I have shown in the diagram how twelve knights (the fewest possible that
will perform the feat) may be placed on the chessboard so that every square
is either occupied or attacked by a knight. Examine every square in turn,
and you will find that this is so. Now, the puzzle in this case is to discover
what is the smallest possible number of knights that is required in order that
every square shall be either occupied or attacked, and every knight
protected by another knight. And how would you arrange them? It will be
found that of the twelve shown in the diagram only four are thus protected
by being a knight's move from another knight.

THE GUARDED CHESSBOARD.
On an ordinary chessboard, 8 by 8, every square can be guarded—that is,
either occupied or attacked—by 5 queens, the fewest possible. There are
exactly 91 fundamentally different arrangements in which no queen attacks
another queen. If every queen must attack (or be protected by) another
queen, there are at fewest 41 arrangements, and I have recorded some 150
ways in which some of the queens are attacked and some not, but this last
case is very difficult to enumerate exactly.
On an ordinary chessboard every square can be guarded by 8 rooks (the
fewest possible) in 40,320 ways, if no rook may attack another rook, but it
is not known how many of these are fundamentally different. (See solution
to No. 295, "The Eight Rooks.") I have not enumerated the ways in which
every rook shall be protected by another rook.
On an ordinary chessboard every square can be guarded by 8 bishops (the
fewest possible), if no bishop may attack another bishop. Ten bishops are
necessary if every bishop is to be protected. (See Nos. 297 and 298,
"Bishops unguarded" and "Bishops guarded.")
On an ordinary chessboard every square can be guarded by 12 knights if all
but 4 are unprotected. But if every knight must be protected, 14 are
necessary. (See No. 319, "The Knight-Guards.")

Dealing with the queen on n2 boards generally, where n is less than 8, the
following results will be of interest:—

1 queen guards 22 board in 1 fundamental way.

1 queen guards 32 board in 1 fundamental way.
2 queens guard 42 board in 3 fundamental ways (protected).

3 queens guard 42 board in 2 fundamental ways (not protected).

3 queens guard 52 board in 37 fundamental ways (protected).

3 queens guard 52 board in 2 fundamental ways (not protected).

3 queens guard 62 board in 1 fundamental way (protected).

4 queens guard 62 board in 17 fundamental ways (not protected).

4 queens guard 72 board in 5 fundamental ways (protected).

4 queens guard 72 board in 1 fundamental way (not protected).

NON-ATTACKING CHESSBOARD ARRANGEMENTS.

We know that n queens may always be placed on a square board of n2
squares (if n be greater than 3) without any queen attacking another queen.
But no general formula for enumerating the number of different ways in
which it may be done has yet been discovered; probably it is
undiscoverable. The known results are as follows:—
Where n = 4 there is 1 fundamental solution and 2 in all.
Where n = 5 there are 2 fundamental solutions and 10 in all.
Where n = 6 there is 1 fundamental solution and 4 in all.
Where n = 7 there are 6 fundamental solutions and 40 in all.
Where n = 8 there are 12 fundamental solutions and 92 in all.
Where n = 9 there are 46 fundamental solutions.
Where n = 10 there are 92 fundamental solutions.
Where n = 11 there are 341 fundamental solutions.

Obviously n rooks may be placed without attack on an n2 board in n! ways,
but how many of these are fundamentally different I have only worked out
in the four cases where n equals 2, 3, 4, and 5. The answers here are
respectively 1, 2, 7, and 23. (See No. 296, "The Four Lions.")

We can place 2n-2 bishops on an n2 board in 2n ways. (See No. 299,
"Bishops in Convocation.") For boards containing 2, 3, 4, 5, 6, 7, 8 squares,

on a side there are respectively 1, 2, 3, 6, 10, 20, 36 fundamentally different
arrangements. Where n is odd there are 2½(n-1) such arrangements, each
giving 4 by reversals and reflections, and 2n-3 - 2½(n-3) giving 8. Where n is
even there are 2½(n-2), each giving 4 by reversals and reflections, and 2n-3 -
2½(n-4), each giving 8.

We can place ½(n2+1) knights on an n2 board without attack, when n is odd,
in 1 fundamental way; and ½n2 knights on an n2 board, when n is even, in 1
fundamental way. In the first case we place all the knights on the same
colour as the central square; in the second case we place them all on black,
or all on white, squares.

THE TWO PIECES PROBLEM.

On a board of n2 squares, two queens, two rooks, two bishops, or two
knights can always be placed, irrespective of attack or not, in ½(n4 - n2)
ways. The following formulæ will show in how many of these ways the two
pieces may be placed with attack and without:—

With Attack. Without Attack.
5n3 - 6n2 + n 3n4 - 10n3 + 9n2 - 2n
2 Queens
3 6
n4 - 2n3 + n2
2 Rooks n3 - n2
2
4n3 - 6n2 + 2n 3n4 - 4n3 + 3n2 - 2n
2 Bishops
6 6
n4 - 9n2 + 24n
2 Knights 4n2 - 12n + 8
2

(See No. 318, "Lion Hunting.")

DYNAMICAL CHESS PUZZLES.
"Push on—keep moving."
THOS. MORTON: Cure for the Heartache.

320.—THE ROOK'S TOUR.

The puzzle is to move the single rook over the whole board, so that it shall
visit every square of the board once, and only once, and end its tour on the
square from which it starts. You have to do this in as few moves as possible,
and unless you are very careful you will take just one move too many. Of
course, a square is regarded equally as "visited" whether you merely pass
over it or make it a stopping-place, and we will not quibble over the point

whether the original square is actually visited twice. We will assume that it
is not.

321.—THE ROOK'S JOURNEY.
This puzzle I call "The Rook's Journey," because the word "tour" (derived
from a turner's wheel) implies that we return to the point from which we set
out, and we do not do this in the present case. We should not be satisfied
with a personally conducted holiday tour that ended by leaving us, say, in
the middle of the Sahara. The rook here makes twenty-one moves, in the
course of which journey it visits every square of the board once and only
once, stopping at the square marked 10 at the end of its tenth move, and
ending at the square marked 21. Two consecutive moves cannot be made in
the same direction—that is to say, you must make a turn after every move.

322.—THE LANGUISHING MAIDEN.

A wicked baron in the good old days imprisoned an innocent maiden in one
of the deepest dungeons beneath the castle moat. It will be seen from our
illustration that there were sixty-three cells in the dungeon, all connected by
open doors, and the maiden was chained in the cell in which she is shown.
Now, a valiant knight, who loved the damsel, succeeded in rescuing her
from the enemy. Having gained an entrance to the dungeon at the point
where he is seen, he succeeded in reaching the maiden after entering every
cell once and only once. Take your pencil and try to trace out such a route.
When you have succeeded, then try to discover a route in twenty-two
straight paths through the cells. It can be done in this number without
entering any cell a second time.

323.—A DUNGEON PUZZLE.

A French prisoner, for his sins (or other people's), was confined in an
underground dungeon containing sixty-four cells, all communicating with
open doorways, as shown in our illustration. In order to reduce the tedium
of his restricted life, he set himself various puzzles, and this is one of them.
Starting from the cell in which he is shown, how could he visit every cell
once, and only once, and make as many turnings as possible? His first
attempt is shown by the dotted track. It will be found that there are as many
as fifty-five straight lines in his path, but after many attempts he improved
upon this. Can you get more than fifty-five? You may end your path in any
cell you like. Try the puzzle with a pencil on chessboard diagrams, or you
may regard them as rooks' moves on a board.

324.—THE LION AND THE MAN.
In a public place in Rome there once stood a prison divided into sixty-four
cells, all open to the sky and all communicating with one another, as shown
in the illustration. The sports that here took place were watched from a high

tower. The favourite game was to place a Christian in one corner cell and a
lion in the diagonally opposite corner and then leave them with all the inner
doors open. The consequent effect was sometimes most laughable. On one
occasion the man was given a sword. He was no coward, and was as
anxious to find the lion as the lion undoubtedly was to find him.

The man visited every cell once and only once in the fewest possible
straight lines until he reached the lion's cell. The lion, curiously enough,
also visited every cell once and only once in the fewest possible straight
lines until he finally reached the man's cell. They started together and went
at the same speed; yet, although they occasionally got glimpses of one
another, they never once met. The puzzle is to show the route that each
happened to take.

325.—AN EPISCOPAL VISITATION.

The white squares on the chessboard represent the parishes of a diocese.
Place the bishop on any square you like, and so contrive that (using the
ordinary bishop's move of chess) he shall visit every one of his parishes in
the fewest possible moves. Of course, all the parishes passed through on
any move are regarded as "visited." You can visit any squares more than
once, but you are not allowed to move twice between the same two
adjoining squares. What are the fewest possible moves? The bishop need
not end his visitation at the parish from which he first set out.

326.—A NEW COUNTER PUZZLE.
Here is a new puzzle with moving counters, or coins, that at first glance
looks as if it must be absurdly simple. But it will be found quite a little
perplexity. I give it in this place for a reason that I will explain when we
come to the next puzzle. Copy the simple diagram, enlarged, on a sheet of
paper; then place two white counters on the points 1 and 2, and two red
counters on 9 and 10, The puzzle is to make the red and white change
places. You may move the counters one at a time in any order you like,
along the lines from point to point, with the only restriction that a red and a
white counter may never stand at once on the same straight line. Thus the
first move can only be from 1 or 2 to 3, or from 9 or 10 to 7.

327.—A NEW BISHOP'S PUZZLE.

This is quite a fascinating little puzzle. Place eight bishops (four black and
four white) on the reduced chessboard, as shown in the illustration. The
problem is to make the black bishops change places with the white ones, no
bishop ever attacking another of the opposite colour. They must move
alternately—first a white, then a black, then a white, and so on. When you
have succeeded in doing it at all, try to find the fewest possible moves.
If you leave out the bishops standing on black squares, and only play on the
white squares, you will discover my last puzzle turned on its side.

328.—THE QUEEN'S TOUR.
The puzzle of making a complete tour of the chessboard with the queen in
the fewest possible moves (in which squares may be visited more than
once) was first given by the late Sam Loyd in his Chess Strategy. But the
solution shown below is the one he gave in American Chess-Nuts in 1868. I
have recorded at least six different solutions in the minimum number of
moves—fourteen—but this one is the best of all, for reasons I will explain.

If you will look at the lettered square you will understand that there are only
ten really differently placed squares on a chessboard—those enclosed by a
dark line—all the others are mere reversals or reflections. For example,
every A is a corner square, and every J a central square. Consequently, as
the solution shown has a turning-point at the enclosed D square, we can
obtain a solution starting from and ending at any square marked D—by just
turning the board about. Now, this scheme will give you a tour starting from
any A, B, C, D, E, F, or H, while no other route that I know can be adapted
to more than five different starting-points. There is no Queen's Tour in
fourteen moves (remember a tour must be re-entrant) that may start from a
G, I, or J. But we can have a non-re-entrant path over the whole board in

fourteen moves, starting from any given square. Hence the following
puzzle:—

Start from the J in the enclosed part of the lettered diagram and visit every
square of the board in fourteen moves, ending wherever you like.

329.—THE STAR PUZZLE.

Put the point of your pencil on one of the white stars and (without ever
lifting your pencil from the paper) strike out all the stars in fourteen
continuous straight strokes, ending at the second white star. Your straight
strokes may be in any direction you like, only every turning must be made
on a star. There is no objection to striking out any star more than once.
In this case, where both your starting and ending squares are fixed
inconveniently, you cannot obtain a solution by breaking a Queen's Tour, or
in any other way by queen moves alone. But you are allowed to use oblique
straight lines—such as from the upper white star direct to a corner star.

330.—THE YACHT RACE.
Now then, ye land-lubbers, hoist your baby-jib-topsails, break out your
spinnakers, ease off your balloon sheets, and get your head-sails set!

Our race consists in starting from the point at which the yacht is lying in the
illustration and touching every one of the sixty-four buoys in fourteen
straight courses, returning in the final tack to the buoy from which we start.
The seventh course must finish at the buoy from which a flag is flying.
This puzzle will call for a lot of skilful seamanship on account of the sharp
angles at which it will occasionally be necessary to tack. The point of a lead
pencil and a good nautical eye are all the outfit that we require.

This is difficult, because of the condition as to the flag-buoy, and because it
is a re-entrant tour. But again we are allowed those oblique lines.

331.—THE SCIENTIFIC SKATER.

It will be seen that this skater has marked on the ice sixty-four points or
stars, and he proposes to start from his present position near the corner and
enter every one of the points in fourteen straight lines. How will he do it?
Of course there is no objection to his passing over any point more than
once, but his last straight stroke must bring him back to the position from
which he started.
It is merely a matter of taking your pencil and starting from the spot on
which the skater's foot is at present resting, and striking out all the stars in
fourteen continuous straight lines, returning to the point from which you set
out.

332.—THE FORTY-NINE STARS.

The puzzle in this case is simply to take your pencil and, starting from one
black star, strike out all the stars in twelve straight strokes, ending at the
other black star. It will be seen that the attempt shown in the illustration
requires fifteen strokes. Can you do it in twelve? Every turning must be
made on a star, and the lines must be parallel to the sides and diagonals of
the square, as shown. In this case we are dealing with a chessboard of
reduced dimensions, but only queen moves (without going outside the
boundary as in the last case) are required.

333.—THE QUEEN'S JOURNEY.

Place the queen on her own square, as shown in the illustration, and then try
to discover the greatest distance that she can travel over the board in five
queen's moves without passing over any square a second time. Mark the
queen's path on the board, and note carefully also that she must never cross
her own track. It seems simple enough, but the reader may find that he has
tripped.

334.—ST. GEORGE AND THE DRAGON.

Here is a little puzzle on a reduced chessboard of forty-nine squares. St.
George wishes to kill the dragon. Killing dragons was a well-known
pastime of his, and, being a knight, it was only natural that he should desire
to perform the feat in a series of knight's moves. Can you show how,
starting from that central square, he may visit once, and only once, every
square of the board in a chain of chess knight's moves, and end by capturing
the dragon on his last move? Of course a variety of different ways are open
to him, so try to discover a route that forms some pretty design when you
have marked each successive leap by a straight line from square to square.

335.—FARMER LAWRENCE'S CORNFIELDS.
One of the most beautiful districts within easy distance of London for a
summer ramble is that part of Buckinghamshire known as the Valley of the
Chess—at least, it was a few years ago, before it was discovered by the
speculative builder. At the beginning of the present century there lived, not
far from Latimers, a worthy but eccentric farmer named Lawrence. One of

his queer notions was that every person who lived near the banks of the
river Chess ought to be in some way acquainted with the noble game of the
same name, and in order to impress this fact on his men and his neighbours
he adopted at times strange terminology. For example, when one of his
ewes presented him with a lamb, he would say that it had "queened a
pawn"; when he put up a new barn against the highway, he called it
"castling on the king's side"; and when he sent a man with a gun to keep his
neighbour's birds off his fields, he spoke of it as "attacking his opponent's
rooks." Everybody in the neighbourhood used to be amused at Farmer
Lawrence's little jokes, and one boy (the wag of the village) who got his
ears pulled by the old gentleman for stealing his "chestnuts" went so far as
to call him "a silly old chess-protector!"
One year he had a large square field divided into forty-nine square plots, as
shown in the illustration. The white squares were sown with wheat and the
black squares with barley. When the harvest time came round he gave
orders that his men were first to cut the corn in the patch marked 1, and that
each successive cutting should be exactly a knight's move from the last one,
the thirteenth cutting being in the patch marked 13, the twenty-fifth in the
patch marked 25, the thirty-seventh in the one marked 37, and the last, or
forty-ninth cutting, in the patch marked 49. This was too much for poor
Hodge, and each day Farmer Lawrence had to go down to the field and
show which piece had to be operated upon. But the problem will perhaps
present no difficulty to my readers.

336.—THE GREYHOUND PUZZLE.
In this puzzle the twenty kennels do not communicate with one another by
doors, but are divided off by a low wall. The solitary occupant is the
greyhound which lives in the kennel in the top left-hand corner. When he is
allowed his liberty he has to obtain it by visiting every kennel once and
only once in a series of knight's moves, ending at the bottom right-hand
corner, which is open to the world. The lines in the above diagram show
one solution. The puzzle is to discover in how many different ways the
greyhound may thus make his exit from his corner kennel.

337.—THE FOUR KANGAROOS.

In introducing a little Commonwealth problem, I must first explain that the
diagram represents the sixty-four fields, all properly fenced off from one
another, of an Australian settlement, though I need hardly say that our kith
and kin "down under" always do set out their land in this methodical and
exact manner. It will be seen that in every one of the four corners is a
kangaroo. Why kangaroos have a marked preference for corner plots has
never been satisfactorily explained, and it would be out of place to discuss
the point here. I should also add that kangaroos, as is well known, always
leap in what we call "knight's moves." In fact, chess players would probably
have adopted the better term "kangaroo's move" had not chess been
invented before kangaroos.
The puzzle is simply this. One morning each kangaroo went for his morning
hop, and in sixteen consecutive knight's leaps visited just fifteen different
fields and jumped back to his corner. No field was visited by more than one
of the kangaroos. The diagram shows how they arranged matters. What you
are asked to do is to show how they might have performed the feat without
any kangaroo ever crossing the horizontal line in the middle of the square
that divides the board into two equal parts.

338.—THE BOARD IN COMPARTMENTS.

We cannot divide the ordinary chessboard into four equal square
compartments, and describe a complete tour, or even path, in each
compartment. But we may divide it into four compartments, as in the
illustration, two containing each twenty squares, and the other two each
twelve squares, and so obtain an interesting puzzle. You are asked to
describe a complete re-entrant tour on this board, starting where you like,
but visiting every square in each successive compartment before passing
into another one, and making the final leap back to the square from which
the knight set out. It is not difficult, but will be found very entertaining and
not uninstructive.
Whether a re-entrant "tour" or a complete knight's "path" is possible or not
on a rectangular board of given dimensions depends not only on its
dimensions, but also on its shape. A tour is obviously not possible on a
board containing an odd number of cells, such as 5 by 5 or 7 by 7, for this
reason: Every successive leap of the knight must be from a white square to
a black and a black to a white alternately. But if there be an odd number of
cells or squares there must be one more square of one colour than of the
other, therefore the path must begin from a square of the colour that is in

excess, and end on a similar colour, and as a knight's move from one colour
to a similar colour is impossible the path cannot be re-entrant. But a perfect
tour may be made on a rectangular board of any dimensions provided the
number of squares be even, and that the number of squares on one side be
not less than 6 and on the other not less than 5. In other words, the smallest
rectangular board on which a re-entrant tour is possible is one that is 6 by 5.
A complete knight's path (not re-entrant) over all the squares of a board is
never possible if there be only two squares on one side; nor is it possible on
a square board of smaller dimensions than 5 by 5. So that on a board 4 by 4
we can neither describe a knight's tour nor a complete knight's path; we
must leave one square unvisited. Yet on a board 4 by 3 (containing four
squares fewer) a complete path may be described in sixteen different ways.
It may interest the reader to discover all these. Every path that starts from
and ends at different squares is here counted as a different solution, and
even reverse routes are called different.

339.—THE FOUR KNIGHTS' TOURS.

I will repeat that if a chessboard be cut into four equal parts, as indicated by
the dark lines in the illustration, it is not possible to perform a knight's tour,
either re-entrant or not, on one of the parts. The best re-entrant attempt is
shown, in which each knight has to trespass twice on other parts. The
puzzle is to cut the board differently into four parts, each of the same size
and shape, so that a re-entrant knight's tour may be made on each part. Cuts
along the dotted lines will not do, as the four central squares of the board
would be either detached or hanging on by a mere thread.

340.—THE CUBIC KNIGHT'S TOUR.
Some few years ago I happened to read somewhere that Abnit
Vandermonde, a clever mathematician, who was born in 1736 and died in
1793, had devoted a good deal of study to the question of knight's tours.
Beyond what may be gathered from a few fragmentary references, I am not
aware of the exact nature or results of his investigations, but one thing
attracted my attention, and that was the statement that he had proposed the

question of a tour of the knight over the six surfaces of a cube, each surface
being a chessboard. Whether he obtained a solution or not I do not know,
but I have never seen one published. So I at once set to work to master this
interesting problem. Perhaps the reader may like to attempt it.

341.—THE FOUR FROGS.

In the illustration we have eight toadstools, with white frogs on 1 and 3 and
black frogs on 6 and 8. The puzzle is to move one frog at a time, in any
order, along one of the straight lines from toadstool to toadstool, until they
have exchanged places, the white frogs being left on 6 and 8 and the black
ones on 1 and 3. If you use four counters on a simple diagram, you will find
this quite easy, but it is a little more puzzling to do it in only seven plays,
any number of successive moves by one frog counting as one play. Of
course, more than one frog cannot be on a toadstool at the same time.

342.—THE MANDARIN'S PUZZLE.
The following puzzle has an added interest from the circumstance that a
correct solution of it secured for a certain young Chinaman the hand of his
charming bride. The wealthiest mandarin within a radius of a hundred miles
of Peking was Hi-Chum-Chop, and his beautiful daughter, Peeky-Bo, had
innumerable admirers. One of her most ardent lovers was Winky-Hi, and
when he asked the old mandarin for his consent to their marriage, Hi-
Chum-Chop presented him with the following puzzle and promised his
consent if the youth brought him the correct answer within a week. Winky-
Hi, following a habit which obtains among certain solvers to this day, gave
it to all his friends, and when he had compared their solutions he handed in
the best one as his own. Luckily it was quite right. The mandarin thereupon
fulfilled his promise. The fatted pup was killed for the wedding feast, and
when Hi-Chum-Chop passed Winky-Hi the liver wing all present knew that
it was a token of eternal goodwill, in accordance with Chinese custom from
time immemorial.
The mandarin had a table divided into twenty-five squares, as shown in the
diagram. On each of twenty-four of these squares was placed a numbered
counter, just as I have indicated. The puzzle is to get the counters in
numerical order by moving them one at a time in what we call "knight's
moves." Counter 1 should be where 16 is, 2 where 11 is, 4 where 13 now is,
and so on. It will be seen that all the counters on shaded squares are in their
proper positions. Of course, two counters may never be on a square at the
same time. Can you perform the feat in the fewest possible moves?

In order to make the manner of moving perfectly clear I will point out that
the first knight's move can only be made by 1 or by 2 or by 10. Supposing 1
moves, then the next move must be by 23, 4, 8, or 21. As there is never
more than one square vacant, the order in which the counters move may be
written out as follows: 1—21—14—18—22, etc. A rough diagram should
be made on a larger scale for practice, and numbered counters or pieces of
cardboard used.

343.—EXERCISE FOR PRISONERS.
The following is the plan of the north wing of a certain gaol, showing the
sixteen cells all communicating by open doorways. Fifteen prisoners were
numbered and arranged in the cells as shown. They were allowed to change
their cells as much as they liked, but if two prisoners were ever in the same
cell together there was a severe punishment promised them.

Now, in order to reduce their growing obesity, and to combine physical
exercise with mental recreation, the prisoners decided, on the suggestion of
one of their number who was interested in knight's tours, to try to form
themselves into a perfect knight's path without breaking the prison
regulations, and leaving the bottom right-hand corner cell vacant, as
originally. The joke of the matter is that the arrangement at which they
arrived was as follows:—

8 3 12 1
11 14 9 6
4 7 2 13
1510 5

The warders failed to detect the important fact that the men could not
possibly get into this position without two of them having been at some
time in the same cell together. Make the attempt with counters on a ruled
diagram, and you will find that this is so. Otherwise the solution is correct

enough, each member being, as required, a knight's move from the
preceding number, and the original corner cell vacant.
The puzzle is to start with the men placed as in the illustration and show
how it might have been done in the fewest moves, while giving a complete
rest to as many prisoners as possible.
As there is never more than one vacant cell for a man to enter, it is only
necessary to write down the numbers of the men in the order in which they
move. It is clear that very few men can be left throughout in their cells
undisturbed, but I will leave the solver to discover just how many, as this is
a very essential part of the puzzle.

344.—THE KENNEL PUZZLE.

A man has twenty-five dog kennels all communicating with each other by
doorways, as shown in the illustration. He wishes to arrange his twenty

dogs so that they shall form a knight's string from dog No. 1 to dog No. 20,
the bottom row of five kennels to be left empty, as at present. This is to be
done by moving one dog at a time into a vacant kennel. The dogs are well
trained to obedience, and may be trusted to remain in the kennels in which
they are placed, except that if two are placed in the same kennel together
they will fight it out to the death. How is the puzzle to be solved in the
fewest possible moves without two dogs ever being together?

345.—THE TWO PAWNS.

Here is a neat little puzzle in counting. In how many different ways may the
two pawns advance to the eighth square? You may move them in any order
you like to form a different sequence. For example, you may move the Q R
P (one or two squares) first, or the K R P first, or one pawn as far as you
like before touching the other. Any sequence is permissible, only in this

puzzle as soon as a pawn reaches the eighth square it is dead, and remains
there unconverted. Can you count the number of different sequences? At
first it will strike you as being very difficult, but I will show that it is really
quite simple when properly attacked.

VARIOUS CHESS PUZZLES.
"Chesse-play is a good and wittie exercise of the minde for some kinde of
men."
Burton's Anatomy of Melancholy.

346.—SETTING THE BOARD.
I have a single chessboard and a single set of chessmen. In how many
different ways may the men be correctly set up for the beginning of a game?
I find that most people slip at a particular point in making the calculation.

347.—COUNTING THE RECTANGLES.
Can you say correctly just how many squares and other rectangles the
chessboard contains? In other words, in how great a number of different
ways is it possible to indicate a square or other rectangle enclosed by lines
that separate the squares of the board?

348.—THE ROOKERY.

The White rooks cannot move outside the little square in which they are
enclosed except on the final move, in giving checkmate. The puzzle is how
to checkmate Black in the fewest possible moves with No. 8 rook, the other
rooks being left in numerical order round the sides of their square with the
break between 1 and 7.

349.—STALEMATE.
Some years ago the puzzle was proposed to construct an imaginary game of
chess, in which White shall be stalemated in the fewest possible moves with
all the thirty-two pieces on the board. Can you build up such a position in
fewer than twenty moves?

350.—THE FORSAKEN KING.

Set up the position shown in the diagram. Then the condition of the puzzle
is—White to play and checkmate in six moves. Notwithstanding the
complexities, I will show how the manner of play may be condensed into
quite a few lines, merely stating here that the first two moves of White
cannot be varied.

351.—THE CRUSADER.
The following is a prize puzzle propounded by me some years ago. Produce
a game of chess which, after sixteen moves, shall leave White with all his
sixteen men on their original squares and Black in possession of his king
alone (not necessarily on his own square). White is then to force mate in
three moves.

352.—IMMOVABLE PAWNS.
Starting from the ordinary arrangement of the pieces as for a game, what is
the smallest possible number of moves necessary in order to arrive at the
following position? The moves for both sides must, of course, be played
strictly in accordance with the rules of the game, though the result will
necessarily be a very weird kind of chess.

353.—THIRTY-SIX MATES.

Place the remaining eight White pieces in such a position that White shall
have the choice of thirty-six different mates on the move. Every move that
checkmates and leaves a different position is a different mate. The pieces
already placed must not be moved.

354.—AN AMAZING DILEMMA.
In a game of chess between Mr. Black and Mr. White, Black was in
difficulties, and as usual was obliged to catch a train. So he proposed that
White should complete the game in his absence on condition that no moves
whatever should be made for Black, but only with the White pieces. Mr.
White accepted, but to his dismay found it utterly impossible to win the
game under such conditions. Try as he would, he could not checkmate his
opponent. On which square did Mr. Black leave his king? The other pieces
are in their proper positions in the diagram. White may leave Black in check
as often as he likes, for it makes no difference, as he can never arrive at a
checkmate position.

355.—CHECKMATE!

Strolling into one of the rooms of a London club, I noticed a position left by
two players who had gone. This position is shown in the diagram. It is
evident that White has checkmated Black. But how did he do it? That is the
puzzle.

356.—QUEER CHESS.
Can you place two White rooks and a White knight on the board so that the
Black king (who must be on one of the four squares in the middle of the
board) shall be in check with no possible move open to him? "In other
words," the reader will say, "the king is to be shown checkmated." Well,
you can use the term if you wish, though I intentionally do not employ it
myself. The mere fact that there is no White king on the board would be a
sufficient reason for my not doing so.

357.—ANCIENT CHINESE PUZZLE.

My next puzzle is supposed to be Chinese, many hundreds of years old, and
never fails to interest. White to play and mate, moving each of the three
pieces once, and once only.

358.—THE SIX PAWNS.
In how many different ways may I place six pawns on the chessboard so
that there shall be an even number of unoccupied squares in every row and
every column? We are not here considering the diagonals at all, and every
different six squares occupied makes a different solution, so we have not to
exclude reversals or reflections.

359.—COUNTER SOLITAIRE.
Here is a little game of solitaire that is quite easy, but not so easy as to be
uninteresting. You can either rule out the squares on a sheet of cardboard or
paper, or you can use a portion of your chessboard. I have shown numbered
counters in the illustration so as to make the solution easy and intelligible to
all, but chess pawns or draughts will serve just as well in practice.

The puzzle is to remove all the counters except one, and this one that is left
must be No. 1. You remove a counter by jumping over another counter to
the next space beyond, if that square is vacant, but you cannot make a leap
in a diagonal direction. The following moves will make the play quite clear:
1-9, 2-10, 1-2, and so on. Here 1 jumps over 9, and you remove 9 from the
board; then 2 jumps over 10, and you remove 10; then 1 jumps over 2, and
you remove 2. Every move is thus a capture, until the last capture of all is
made by No. 1.

360.—CHESSBOARD SOLITAIRE.

Here is an extension of the last game of solitaire. All you need is a
chessboard and the thirty-two pieces, or the same number of draughts or
counters. In the illustration numbered counters are used. The puzzle is to
remove all the counters except two, and these two must have originally
been on the same side of the board; that is, the two left must either belong
to the group 1 to 16 or to the other group, 17 to 32. You remove a counter
by jumping over it with another counter to the next square beyond, if that
square is vacant, but you cannot make a leap in a diagonal direction. The
following moves will make the play quite clear: 3-11, 4-12, 3-4, 13-3. Here
3 jumps over 11, and you remove 11; 4 jumps over 12, and you remove 12;
and so on. It will be found a fascinating little game of patience, and the
solution requires the exercise of some ingenuity.

361.—THE MONSTROSITY.
One Christmas Eve I was travelling by rail to a little place in one of the
southern counties. The compartment was very full, and the passengers were

wedged in very tightly. My neighbour in one of the corner seats was closely
studying a position set up on one of those little folding chessboards that can
be carried conveniently in the pocket, and I could scarcely avoid looking at
it myself. Here is the position:—

My fellow-passenger suddenly turned his head and caught the look of
bewilderment on my face.
"Do you play chess?" he asked.
"Yes, a little. What is that? A problem?"
"Problem? No; a game."
"Impossible!" I exclaimed rather rudely. "The position is a perfect
monstrosity!"
He took from his pocket a postcard and handed it to me. It bore an address
at one side and on the other the words "43. K to Kt 8."

"It is a correspondence game." he exclaimed. "That is my friend's last move,
and I am considering my reply."
"But you really must excuse me; the position seems utterly impossible.
How on earth, for example—"
"Ah!" he broke in smilingly. "I see; you are a beginner; you play to win."
"Of course you wouldn't play to lose or draw!"
He laughed aloud.
"You have much to learn. My friend and myself do not play for results of
that antiquated kind. We seek in chess the wonderful, the whimsical, the
weird. Did you ever see a position like that?"
I inwardly congratulated myself that I never had.
"That position, sir, materializes the sinuous evolvements and syncretic,
synthetic, and synchronous concatenations of two cerebral individualities. It
is the product of an amphoteric and intercalatory interchange of—"
"Have you seen the evening paper, sir?" interrupted the man opposite,
holding out a newspaper. I noticed on the margin beside his thumb some
pencilled writing. Thanking him, I took the paper and read—"Insane, but
quite harmless. He is in my charge."
After that I let the poor fellow run on in his wild way until both got out at
the next station.
But that queer position became fixed indelibly in my mind, with Black's last
move 43. K to Kt 8; and a short time afterwards I found it actually possible
to arrive at such a position in forty-three moves. Can the reader construct
such a sequence? How did White get his rooks and king's bishop into their
present positions, considering Black can never have moved his king's
bishop? No odds were given, and every move was perfectly legitimate.

MEASURING, WEIGHING, AND PACKING
PUZZLES.
"Measure still for measure."
Measure for Measure, v. 1.
Apparently the first printed puzzle involving the measuring of a given
quantity of liquid by pouring from one vessel to others of known capacity
was that propounded by Niccola Fontana, better known as "Tartaglia" (the
stammerer), 1500-1559. It consists in dividing 24 oz. of valuable balsam
into three equal parts, the only measures available being vessels holding 5,
11, and 13 ounces respectively. There are many different solutions to this
puzzle in six manipulations, or pourings from one vessel to another. Bachet
de Méziriac reprinted this and other of Tartaglia's puzzles in his Problèmes
plaisans et délectables (1612). It is the general opinion that puzzles of this
class can only be solved by trial, but I think formulæ can be constructed for
the solution generally of certain related cases. It is a practically unexplored
field for investigation.
The classic weighing problem is, of course, that proposed by Bachet. It
entails the determination of the least number of weights that would serve to
weigh any integral number of pounds from 1 lb. to 40 lbs. inclusive, when
we are allowed to put a weight in either of the two pans. The answer is 1, 3,
9, and 27 lbs. Tartaglia had previously propounded the same puzzle with the
condition that the weights may only be placed in one pan. The answer in
that case is 1, 2, 4, 8, 16, 32 lbs. Major MacMahon has solved the problem
quite generally. A full account will be found in Ball's Mathematical
Recreations (5th edition).
Packing puzzles, in which we are required to pack a maximum number of
articles of given dimensions into a box of known dimensions, are, I believe,
of quite recent introduction. At least I cannot recall any example in the
books of the old writers. One would rather expect to find in the toy shops
the idea presented as a mechanical puzzle, but I do not think I have ever
seen such a thing. The nearest approach to it would appear to be the puzzles

of the jig-saw character, where there is only one depth of the pieces to be
adjusted.

362.—THE WASSAIL BOWL.
One Christmas Eve three Weary Willies came into possession of what was
to them a veritable wassail bowl, in the form of a small barrel, containing
exactly six quarts of fine ale. One of the men possessed a five-pint jug and
another a three-pint jug, and the problem for them was to divide the liquor
equally amongst them without waste. Of course, they are not to use any
other vessels or measures. If you can show how it was to be done at all,
then try to find the way that requires the fewest possible manipulations,
every separate pouring from one vessel to another, or down a man's throat,
counting as a manipulation.

363.—THE DOCTOR'S QUERY.
"A curious little point occurred to me in my dispensary this morning," said a
doctor. "I had a bottle containing ten ounces of spirits of wine, and another
bottle containing ten ounces of water. I poured a quarter of an ounce of
spirits into the water and shook them up together. The mixture was then
clearly forty to one. Then I poured back a quarter-ounce of the mixture, so
that the two bottles should again each contain the same quantity of fluid.
What proportion of spirits to water did the spirits of wine bottle then
contain?"

364.—THE BARREL PUZZLE.
The men in the illustration are disputing over the liquid contents of a barrel.
What the particular liquid is it is impossible to say, for we are unable to

look into the barrel; so we will call it water. One man says that the barrel is
more than half full, while the other insists that it is not half full. What is
their easiest way of settling the point? It is not necessary to use stick, string,
or implement of any kind for measuring. I give this merely as one of the
simplest possible examples of the value of ordinary sagacity in the solving
of puzzles. What are apparently very difficult problems may frequently be
solved in a similarly easy manner if we only use a little common sense.

365.—NEW MEASURING PUZZLE.
Here is a new poser in measuring liquids that will be found interesting. A
man has two ten-quart vessels full of wine, and a five-quart and a four-quart
measure. He wants to put exactly three quarts into each of the two
measures. How is he to do it? And how many manipulations (pourings from

one vessel to another) do you require? Of course, waste of wine, tilting, and
other tricks are not allowed.

366.—THE HONEST DAIRYMAN.
An honest dairyman in preparing his milk for public consumption employed
a can marked B, containing milk, and a can marked A, containing water.
From can A he poured enough to double the contents of can B. Then he
poured from can B into can A enough to double its contents. Then he finally
poured from can A into can B until their contents were exactly equal. After
these operations he would send the can A to London, and the puzzle is to
discover what are the relative proportions of milk and water that he
provides for the Londoners' breakfast-tables. Do they get equal proportions
of milk and water—or two parts of milk and one of water—or what? It is an
interesting question, though, curiously enough, we are not told how much
milk or water he puts into the cans at the start of his operations.

367.—WINE AND WATER.
Mr. Goodfellow has adopted a capital idea of late. When he gives a little
dinner party and the time arrives to smoke, after the departure of the ladies,
he sometimes finds that the conversation is apt to become too political, too
personal, too slow, or too scandalous. Then he always manages to introduce
to the company some new poser that he has secreted up his sleeve for the
occasion. This invariably results in no end of interesting discussion and
debate, and puts everybody in a good humour.
Here is a little puzzle that he propounded the other night, and it is
extraordinary how the company differed in their answers. He filled a wine-
glass half full of wine, and another glass twice the size one-third full of
wine. Then he filled up each glass with water and emptied the contents of
both into a tumbler. "Now," he said, "what part of the mixture is wine and
what part water?" Can you give the correct answer?

368.—THE KEG OF WINE.
Here is a curious little problem. A man had a ten-gallon keg full of wine and
a jug. One day he drew off a jugful of wine and filled up the keg with water.
Later on, when the wine and water had got thoroughly mixed, he drew off
another jugful and again filled up the keg with water. It was then found that
the keg contained equal proportions of wine and water. Can you find from
these facts the capacity of the jug?

369.—MIXING THE TEA.
"Mrs. Spooner called this morning," said the honest grocer to his assistant.
"She wants twenty pounds of tea at 2s. 4½d. per lb. Of course we have a
good 2s. 6d. tea, a slightly inferior at 2s. 3d., and a cheap Indian at 1s. 9d.,
but she is very particular always about her prices."
"What do you propose to do?" asked the innocent assistant.
"Do?" exclaimed the grocer. "Why, just mix up the three teas in different
proportions so that the twenty pounds will work out fairly at the lady's
price. Only don't put in more of the best tea than you can help, as we make
less profit on that, and of course you will use only our complete pound
packets. Don't do any weighing."
How was the poor fellow to mix the three teas? Could you have shown him
how to do it?

370.—A PACKING PUZZLE.
As we all know by experience, considerable ingenuity is often required in
packing articles into a box if space is not to be unduly wasted. A man once
told me that he had a large number of iron balls, all exactly two inches in

diameter, and he wished to pack as many of these as possible into a
rectangular box 249/10 inches long, 224/5 inches wide, and 14 inches deep.
Now, what is the greatest number of the balls that he could pack into that
box?

371.—GOLD PACKING IN RUSSIA.
The editor of the Times newspaper was invited by a high Russian official to
inspect the gold stored in reserve at St. Petersburg, in order that he might
satisfy himself that it was not another "Humbert safe." He replied that it
would be of no use whatever, for although the gold might appear to be
there, he would be quite unable from a mere inspection to declare that what
he saw was really gold. A correspondent of the Daily Mail thereupon took
up the challenge, but, although he was greatly impressed by what he saw, he
was compelled to confess his incompetence (without emptying and
counting the contents of every box and sack, and assaying every piece of
gold) to give any assurance on the subject. In presenting the following little
puzzle, I wish it to be also understood that I do not guarantee the real
existence of the gold, and the point is not at all material to our purpose.
Moreover, if the reader says that gold is not usually "put up" in slabs of the
dimensions that I give, I can only claim problematic licence.
Russian officials were engaged in packing 800 gold slabs, each measuring
12½ inches long, 11 inches wide, and 1 inch deep. What are the interior
dimensions of a box of equal length and width, and necessary depth, that
will exactly contain them without any space being left over? Not more than
twelve slabs may be laid on edge, according to the rules of the government.
It is an interesting little problem in packing, and not at all difficult.

372.—THE BARRELS OF HONEY.

Once upon a time there was an aged merchant of Bagdad who was much
respected by all who knew him. He had three sons, and it was a rule of his
life to treat them all exactly alike. Whenever one received a present, the
other two were each given one of equal value. One day this worthy man fell
sick and died, bequeathing all his possessions to his three sons in equal
shares.
The only difficulty that arose was over the stock of honey. There were
exactly twenty-one barrels. The old man had left instructions that not only
should every son receive an equal quantity of honey, but should receive
exactly the same number of barrels, and that no honey should be transferred
from barrel to barrel on account of the waste involved. Now, as seven of
these barrels were full of honey, seven were half-full, and seven were
empty, this was found to be quite a puzzle, especially as each brother
objected to taking more than four barrels of, the same description—full,
half-full, or empty. Can you show how they succeeded in making a correct
division of the property?

CROSSING RIVER PROBLEMS
"My boat is on the shore."
BYRON.
This is another mediæval class of puzzles. Probably the earliest example
was by Abbot Alcuin, who was born in Yorkshire in 735 and died at Tours
in 804. And everybody knows the story of the man with the wolf, goat, and
basket of cabbages whose boat would only take one of the three at a time
with the man himself. His difficulties arose from his being unable to leave
the wolf alone with the goat, or the goat alone with the cabbages. These
puzzles were considered by Tartaglia and Bachet, and have been later
investigated by Lucas, De Fonteney, Delannoy, Tarry, and others. In the
puzzles I give there will be found one or two new conditions which add to
the complexity somewhat. I also include a pulley problem that practically
involves the same principles.

373.—CROSSING THE STREAM.
During a country ramble Mr. and Mrs. Softleigh found themselves in a
pretty little dilemma. They had to cross a stream in a small boat which was
capable of carrying only 150 lbs. weight. But Mr. Softleigh and his wife
each weighed exactly 150 lbs., and each of their sons weighed 75 lbs. And
then there was the dog, who could not be induced on any terms to swim. On
the principle of "ladies first," they at once sent Mrs. Softleigh over; but this
was a stupid oversight, because she had to come back again with the boat,
so nothing was gained by that operation. How did they all succeed in
getting across? The reader will find it much easier than the Softleigh family
did, for their greatest enemy could not have truthfully called them a brilliant
quartette—while the dog was a perfect fool.

374.—CROSSING THE RIVER AXE.

Many years ago, in the days of the smuggler known as "Rob Roy of the
West," a piratical band buried on the coast of South Devon a quantity of
treasure which was, of course, abandoned by them in the usual inexplicable
way. Some time afterwards its whereabouts was discovered by three
countrymen, who visited the spot one night and divided the spoil between
them, Giles taking treasure to the value of £800, Jasper £500 worth, and
Timothy £300 worth. In returning they had to cross the river Axe at a point
where they had left a small boat in readiness. Here, however, was a
difficulty they had not anticipated. The boat would only carry two men, or
one man and a sack, and they had so little confidence in one another that no
person could be left alone on the land or in the boat with more than his
share of the spoil, though two persons (being a check on each other) might
be left with more than their shares. The puzzle is to show how they got over
the river in the fewest possible crossings, taking their treasure with them.

No tricks, such as ropes, "flying bridges," currents, swimming, or similar
dodges, may be employed.

375.—FIVE JEALOUS HUSBANDS.
During certain local floods five married couples found themselves
surrounded by water, and had to escape from their unpleasant position in a
boat that would only hold three persons at a time. Every husband was so
jealous that he would not allow his wife to be in the boat or on either bank
with another man (or with other men) unless he was himself present. Show
the quickest way of getting these five men and their wives across into
safety.
Call the men A, B, C, D, E, and their respective wives a, b, c, d, e. To go
over and return counts as two crossings. No tricks such as ropes, swimming,
currents, etc., are permitted.

376.—THE FOUR ELOPEMENTS.
Colonel B—— was a widower of a very taciturn disposition. His treatment
of his four daughters was unusually severe, almost cruel, and they not
unnaturally felt disposed to resent it. Being charming girls with every virtue
and many accomplishments, it is not surprising that each had a fond
admirer. But the father forbade the young men to call at his house,
intercepted all letters, and placed his daughters under stricter supervision
than ever. But love, which scorns locks and keys and garden walls, was
equal to the occasion, and the four youths conspired together and planned a
general elopement.
At the foot of the tennis lawn at the bottom of the garden ran the silver
Thames, and one night, after the four girls had been safely conducted from
a dormitory window to terra firma, they all crept softly down to the bank of
the river, where a small boat belonging to the Colonel was moored. With

this they proposed to cross to the opposite side and make their way to a lane
where conveyances were waiting to carry them in their flight. Alas! here at
the water's brink their difficulties already began.
The young men were so extremely jealous that not one of them would allow
his prospective bride to remain at any time in the company of another man,
or men, unless he himself were present also. Now, the boat would only hold
two persons, though it could, of course, be rowed by one, and it seemed
impossible that the four couples would ever get across. But midway in the
stream was a small island, and this seemed to present a way out of the
difficulty, because a person or persons could be left there while the boat was
rowed back or to the opposite shore. If they had been prepared for their
difficulty they could have easily worked out a solution to the little poser at
any other time. But they were now so hurried and excited in their flight that
the confusion they soon got into was exceedingly amusing—or would have
been to any one except themselves.
As a consequence they took twice as long and crossed the river twice as
often as was really necessary. Meanwhile, the Colonel, who was a very light
sleeper, thought he heard a splash of oars. He quickly raised the alarm
among his household, and the young ladies were found to be missing.
Somebody was sent to the police-station, and a number of officers soon
aided in the pursuit of the fugitives, who, in consequence of that delay in
crossing the river, were quickly overtaken. The four girls returned sadly to
their homes, and afterwards broke off their engagements in disgust.
For a considerable time it was a mystery how the party of eight managed to
cross the river in that little boat without any girl being ever left with a man,
unless her betrothed was also present. The favourite method is to take eight
counters or pieces of cardboard and mark them A, B, C, D, a, b, c, d, to
represent the four men and their prospective brides, and carry them from
one side of a table to the other in a matchbox (to represent the boat), a
penny being placed in the middle of the table as the island.
Readers are now asked to find the quickest method of getting the party
across the river. How many passages are necessary from land to land? By
"land" is understood either shore or island. Though the boat would not
necessarily call at the island every time of crossing, the possibility of its

doing so must be provided for. For example, it would not do for a man to be
alone in the boat (though it were understood that he intended merely to
cross from one bank to the opposite one) if there happened to be a girl alone
on the island other than the one to whom he was engaged.

377.—STEALING THE CASTLE TREASURE.
The ingenious manner in which a box of treasure, consisting principally of
jewels and precious stones, was stolen from Gloomhurst Castle has been
handed down as a tradition in the De Gourney family. The thieves consisted
of a man, a youth, and a small boy, whose only mode of escape with the box
of treasure was by means of a high window. Outside the window was fixed
a pulley, over which ran a rope with a basket at each end. When one basket
was on the ground the other was at the window. The rope was so disposed
that the persons in the basket could neither help themselves by means of it
nor receive help from others. In short, the only way the baskets could be
used was by placing a heavier weight in one than in the other.
Now, the man weighed 195 lbs., the youth 105 lbs., the boy 90 lbs., and the
box of treasure 75 lbs. The weight in the descending basket could not
exceed that in the other by more than 15 lbs. without causing a descent so
rapid as to be most dangerous to a human being, though it would not injure
the stolen property. Only two persons, or one person and the treasure, could
be placed in the same basket at one time. How did they all manage to
escape and take the box of treasure with them?
The puzzle is to find the shortest way of performing the feat, which in itself
is not difficult. Remember, a person cannot help himself by hanging on to
the rope, the only way being to go down "with a bump," with the weight in
the other basket as a counterpoise.

PROBLEMS CONCERNING GAMES.
"The little pleasure of the game."
MATTHEW PRIOR.
Every game lends itself to the propounding of a variety of puzzles. They
can be made, as we have seen, out of the chessboard and the peculiar moves
of the chess pieces. I will now give just a few examples of puzzles with
playing cards and dominoes, and also go out of doors and consider one or
two little posers in the cricket field, at the football match, and the horse race
and motor-car race.

378.—DOMINOES IN PROGRESSION.

It will be seen that I have played six dominoes, in the illustration, in
accordance with the ordinary rules of the game, 4 against 4, 1 against 1, and
so on, and yet the sum of the spots on the successive dominoes, 4, 5, 6, 7, 8,
9, are in arithmetical progression; that is, the numbers taken in order have a
common difference of 1. In how many different ways may we play six
dominoes, from an ordinary box of twenty-eight, so that the numbers on
them may lie in arithmetical progression? We must always play from left to

right, and numbers in decreasing arithmetical progression (such as 9, 8, 7, 6,
5, 4) are not admissible.

379.—THE FIVE DOMINOES.

Here is a new little puzzle that is not difficult, but will probably be found
entertaining by my readers. It will be seen that the five dominoes are so
arranged in proper sequence (that is, with 1 against 1, 2 against 2, and so
on), that the total number of pips on the two end dominoes is five, and the
sum of the pips on the three dominoes in the middle is also five. There are
just three other arrangements giving five for the additions. They are: —

(1—0) (0—0) (0—2) (2—1) (1—3)
(4—0) (0—0) (0—2) (2—1) (1—0)
(2—0) (0—0) (0—1) (1—3) (3—0)

Now, how many similar arrangements are there of five dominoes that shall
give six instead of five in the two additions?

380.—THE DOMINO FRAME PUZZLE.

It will be seen in the illustration that the full set of twenty-eight dominoes is
arranged in the form of a square frame, with 6 against 6, 2 against 2, blank
against blank, and so on, as in the game. It will be found that the pips in the
top row and left-hand column both add up 44. The pips in the other two
sides sum to 59 and 32 respectively. The puzzle is to rearrange the
dominoes in the same form so that all of the four sides shall sum to 44.
Remember that the dominoes must be correctly placed one against another
as in the game.

381.—THE CARD FRAME PUZZLE.
In the illustration we have a frame constructed from the ten playing cards,
ace to ten of diamonds. The children who made it wanted the pips on all
four sides to add up alike, but they failed in their attempt and gave it up as
impossible. It will be seen that the pips in the top row, the bottom row, and
the left-hand side all add up 14, but the right-hand side sums to 23. Now,
what they were trying to do is quite possible. Can you rearrange the ten

cards in the same formation so that all four sides shall add up alike? Of
course they need not add up 14, but any number you choose to select.

382.—THE CROSS OF CARDS.

In this case we use only nine cards—the ace to nine of diamonds. The
puzzle is to arrange them in the form of a cross, exactly in the way shown in
the illustration, so that the pips in the vertical bar and in the horizontal bar
add up alike. In the example given it will be found that both directions add
up 23. What I want to know is, how many different ways are there of
rearranging the cards in order to bring about this result? It will be seen that,
without affecting the solution, we may exchange the 5 with the 6, the 5 with
the 7, the 8 with the 3, and so on. Also we may make the horizontal and the
vertical bars change places. But such obvious manipulations as these are not
to be regarded as different solutions. They are all mere variations of one
fundamental solution. Now, how many of these fundamentally different
solutions are there? The pips need not, of course, always add up 23.

383.—THE "T" CARD PUZZLE.

An entertaining little puzzle with cards is to take the nine cards of a suit,
from ace to nine inclusive, and arrange them in the form of the letter "T," as
shown in the illustration, so that the pips in the horizontal line shall count
the same as those in the column. In the example given they add up twenty-
three both ways. Now, it is quite easy to get a single correct arrangement.
The puzzle is to discover in just how many different ways it may be done.
Though the number is high, the solution is not really difficult if we attack
the puzzle in the right manner. The reverse way obtained by reflecting the
illustration in a mirror we will not count as different, but all other changes
in the relative positions of the cards will here count. How many different
ways are there?

384.—CARD TRIANGLES.
Here you pick out the nine cards, ace to nine of diamonds, and arrange them
in the form of a triangle, exactly as shown in the illustration, so that the pips
add up the same on the three sides. In the example given it will be seen that
they sum to 20 on each side, but the particular number is of no importance

so long as it is the same on all three sides. The puzzle is to find out in just
how many different ways this can be done.
If you simply turn the cards round so that one of the other two sides is
nearest to you this will not count as different, for the order will be the same.
Also, if you make the 4, 9, 5 change places with the 7, 3, 8, and at the same
time exchange the 1 and the 6, it will not be different. But if you only
change the 1 and the 6 it will be different, because the order round the
triangle is not the same. This explanation will prevent any doubt arising as
to the conditions.

385.—"STRAND" PATIENCE.
The idea for this came to me when considering the game of Patience that I
gave in the Strand Magazine for December, 1910, which has been reprinted
in Ernest Bergholt's Second Book of Patience Games, under the new name
of "King Albert."

Make two piles of cards as follows: 9 D, 8 S, 7 D, 6 S, 5 D, 4 S, 3 D, 2 S, 1
D, and 9 H, 8 C, 7 H, 6 C, 5 H, 4 C, 3 H, 2 C, 1 H, with the 9 of diamonds
at the bottom of one pile and the 9 of hearts at the bottom of the other. The
point is to exchange the spades with the clubs, so that the diamonds and
clubs are still in numerical order in one pile and the hearts and spades in the
other. There are four vacant spaces in addition to the two spaces occupied
by the piles, and any card may be laid on a space, but a card can only be
laid on another of the next higher value—an ace on a two, a two on a three,
and so on. Patience is required to discover the shortest way of doing this.
When there are four vacant spaces you can pile four cards in seven moves,
with only three spaces you can pile them in nine moves, and with two
spaces you cannot pile more than two cards. When you have a grasp of
these and similar facts you will be able to remove a number of cards bodily
and write down 7, 9, or whatever the number of moves may be. The gradual
shortening of play is fascinating, and first attempts are surprisingly lengthy.

386.—A TRICK WITH DICE.

Here is a neat little trick with three dice. I ask you to throw the dice without
my seeing them. Then I tell you to multiply the points of the first die by 2
and add 5; then multiply the result by 5 and add the points of the second
die; then multiply the result by 10 and add the points of the third die. You
then give me the total, and I can at once tell you the points thrown with the
three dice. How do I do it? As an example, if you threw 1, 3, and 6, as in the
illustration, the result you would give me would be 386, from which I could
at once say what you had thrown.

387.—THE VILLAGE CRICKET MATCH.
In a cricket match, Dingley Dell v. All Muggleton, the latter had the first
innings. Mr. Dumkins and Mr. Podder were at the wickets, when the wary
Dumkins made a splendid late cut, and Mr. Podder called on him to run.
Four runs were apparently completed, but the vigilant umpires at each end
called, "three short," making six short runs in all. What number did Mr.
Dumkins score? When Dingley Dell took their turn at the wickets their
champions were Mr. Luffey and Mr. Struggles. The latter made a
magnificent off-drive, and invited his colleague to "come along," with the
result that the observant spectators applauded them for what was supposed
to have been three sharp runs. But the umpires declared that there had been
two short runs at each end—four in all. To what extent, if any, did this
manœuvre increase Mr. Struggles's total?

388.—SLOW CRICKET.
In the recent county match between Wessex and Nincomshire the former
team were at the wickets all day, the last man being put out a few minutes
before the time for drawing stumps. The play was so slow that most of the
spectators were fast asleep, and, on being awakened by one of the officials
clearing the ground, we learnt that two men had been put out leg-before-
wicket for a combined score of 19 runs; four men were caught for a
combined score or 17 runs; one man was run out for a duck's egg; and the
others were all bowled for 3 runs each. There were no extras. We were not
told which of the men was the captain, but he made exactly 15 more than
the average of his team. What was the captain's score?

389.—THE FOOTBALL PLAYERS.

"It is a glorious game!" an enthusiast was heard to exclaim. "At the close of
last season, of the footballers of my acquaintance four had broken their left
arm, five had broken their right arm, two had the right arm sound, and three
had sound left arms." Can you discover from that statement what is the
smallest number of players that the speaker could be acquainted with?
It does not at all follow that there were as many as fourteen men, because,
for example, two of the men who had broken the left arm might also be the
two who had sound right arms.

390.—THE HORSE-RACE PUZZLE.
There are no morals in puzzles. When we are solving the old puzzle of the
captain who, having to throw half his crew overboard in a storm, arranged
to draw lots, but so placed the men that only the Turks were sacrificed, and
all the Christians left on board, we do not stop to discuss the questionable
morality of the proceeding. And when we are dealing with a measuring
problem, in which certain thirsty pilgrims are to make an equitable division
of a barrel of beer, we do not object that, as total abstainers, it is against our
conscience to have anything to do with intoxicating liquor. Therefore I
make no apology for introducing a puzzle that deals with betting.
Three horses—Acorn, Bluebottle, and Capsule—start in a race. The odds
are 4 to 1, Acorn; 3 to 1, Bluebottle; 2 to 1, Capsule. Now, how much must
I invest on each horse in order to win £13, no matter which horse comes in
first? Supposing, as an example, that I betted £5 on each horse. Then, if
Acorn won, I should receive £20 (four times £5), and have to pay £5 each
for the other two horses; thereby winning £10. But it will be found that if
Bluebottle was first I should only win £5, and if Capsule won I should gain
nothing and lose nothing. This will make the question perfectly clear to the
novice, who, like myself, is not interested in the calling of the fraternity
who profess to be engaged in the noble task of "improving the breed of
horses."

391.—THE MOTOR-CAR RACE.
Sometimes a quite simple statement of fact, if worded in an unfamiliar
manner, will cause considerable perplexity. Here is an example, and it will
doubtless puzzle some of my more youthful readers just a little. I happened
to be at a motor-car race at Brooklands, when one spectator said to another,
while a number of cars were whirling round and round the circular track:—
"There's Gogglesmith—that man in the white car!"
"Yes, I see," was the reply; "but how many cars are running in this race?"
Then came this curious rejoinder:—
"One-third of the cars in front of Gogglesmith added to three-quarters of
those behind him will give you the answer."
Now, can you tell how many cars were running in the race?

PUZZLE GAMES.
"He that is beaten may be said
To lie in honour's truckle bed."
HUDIBRAS.
It may be said generally that a game is a contest of skill for two or more
persons, into which we enter either for amusement or to win a prize. A
puzzle is something to be done or solved by the individual. For example, if
it were possible for us so to master the complexities of the game of chess
that we could be assured of always winning with the first or second move,
as the case might be, or of always drawing, then it would cease to be a
game and would become a puzzle. Of course among the young and
uninformed, when the correct winning play is not understood, a puzzle may
well make a very good game. Thus there is no doubt children will continue
to play "Noughts and Crosses," though I have shown (No. 109, "Canterbury
Puzzles") that between two players who both thoroughly understand the
play, every game should be drawn. Neither player could ever win except
through the blundering of his opponent. But I am writing from the point of
view of the student of these things.
The examples that I give in this class are apparently games, but, since I
show in every case how one player may win if he only play correctly, they
are in reality puzzles. Their interest, therefore, lies in attempting to discover
the leading method of play.

392.—THE PEBBLE GAME.
Here is an interesting little puzzle game that I used to play with an
acquaintance on the beach at Slocomb-on-Sea. Two players place an odd
number of pebbles, we will say fifteen, between them. Then each takes in
turn one, two, or three pebbles (as he chooses), and the winner is the one
who gets the odd number. Thus, if you get seven and your opponent eight,

you win. If you get six and he gets nine, he wins. Ought the first or second
player to win, and how? When you have settled the question with fifteen
pebbles try again with, say, thirteen.

393.—THE TWO ROOKS.
This is a puzzle game for two players. Each player has a single rook. The
first player places his rook on any square of the board that he may choose to
select, and then the second player does the same. They now play in turn, the
point of each play being to capture the opponent's rook. But in this game
you cannot play through a line of attack without being captured. That is to
say, if in the diagram it is Black's turn to play, he cannot move his rook to
his king's knight's square, or to his king's rook's square, because he would
enter the "line of fire" when passing his king's bishop's square. For the same
reason he cannot move to his queen's rook's seventh or eighth squares. Now,
the game can never end in a draw. Sooner or later one of the rooks must
fall, unless, of course, both players commit the absurdity of not trying to
win. The trick of winning is ridiculously simple when you know it. Can you
solve the puzzle?

394.—PUSS IN THE CORNER.

This variation of the last puzzle is also played by two persons. One puts a
counter on No. 6, and the other puts one on No. 55, and they play
alternately by removing the counter to any other number in a line. If your
opponent moves at any time on to one of the lines you occupy, or even
crosses one of your lines, you immediately capture him and win. We will
take an illustrative game.
A moves from 55 to 52; B moves from 6 to 13; A advances to 23; B goes to
15; A retreats to 26; B retreats to 13; A advances to 21; B retreats to 2; A
advances to 7; B goes to 3; A moves to 6; B must now go to 4; A establishes
himself at 11, and B must be captured next move because he is compelled
to cross a line on which A stands. Play this over and you will understand the
game directly. Now, the puzzle part of the game is this: Which player
should win, and how many moves are necessary?

395.—A WAR PUZZLE GAME.

Here is another puzzle game. One player, representing the British general,
places a counter at B, and the other player, representing the enemy, places
his counter at E. The Britisher makes the first advance along one of the
roads to the next town, then the enemy moves to one of his nearest towns,
and so on in turns, until the British general gets into the same town as the
enemy and captures him. Although each must always move along a road to
the next town only, and the second player may do his utmost to avoid
capture, the British general (as we should suppose, from the analogy of real
life) must infallibly win. But how? That is the question.

396.—A MATCH MYSTERY.
Here is a little game that is childishly simple in its conditions. But it is well
worth investigation.
Mr. Stubbs pulled a small table between himself and his friend, Mr. Wilson,
and took a box of matches, from which he counted out thirty.

"Here are thirty matches," he said. "I divide them into three unequal heaps.
Let me see. We have 14, 11, and 5, as it happens. Now, the two players
draw alternately any number from any one heap, and he who draws the last
match loses the game. That's all! I will play with you, Wilson. I have
formed the heaps, so you have the first draw."
"As I can draw any number," Mr. Wilson said, "suppose I exhibit my usual
moderation and take all the 14 heap."
"That is the worst you could do, for it loses right away. I take 6 from the 11,
leaving two equal heaps of 5, and to leave two equal heaps is a certain win
(with the single exception of 1, 1), because whatever you do in one heap I
can repeat in the other. If you leave 4 in one heap, I leave 4 in the other. If
you then leave 2 in one heap, I leave 2 in the other. If you leave only 1 in
one heap, then I take all the other heap. If you take all one heap, I take all
but one in the other. No, you must never leave two heaps, unless they are
equal heaps and more than 1, 1. Let's begin again."
"Very well, then," said Mr. Wilson. "I will take 6 from the 14, and leave you
8, 11, 5."
Mr. Stubbs then left 8, 11, 3; Mr. Wilson, 8, 5, 3; Mr. Stubbs, 6, 5, 3; Mr.
Wilson,4, 5, 3; Mr. Stubbs, 4, 5, 1; Mr. Wilson, 4, 3, 1; Mr. Stubbs, 2, 3, 1;
Mr. Wilson, 2, 1, 1; which Mr. Stubbs reduced to 1, 1, 1.
"It is now quite clear that I must win," said Mr. Stubbs, because you must
take 1, and then I take 1, leaving you the last match. You never had a
chance. There are just thirteen different ways in which the matches may be
grouped at the start for a certain win. In fact, the groups selected, 14, 11, 5,
are a certain win, because for whatever your opponent may play there is
another winning group you can secure, and so on and on down to the last
match."

397.—THE MONTENEGRIN DICE GAME.
It is said that the inhabitants of Montenegro have a little dice game that is
both ingenious and well worth investigation. The two players first select

two different pairs of odd numbers (always higher than 3) and then
alternately toss three dice. Whichever first throws the dice so that they add
up to one of his selected numbers wins. If they are both successful in two
successive throws it is a draw and they try again. For example, one player
may select 7 and 15 and the other 5 and 13. Then if the first player throws
so that the three dice add up 7 or 15 he wins, unless the second man gets
either 5 or 13 on his throw.
The puzzle is to discover which two pairs of numbers should be selected in
order to give both players an exactly even chance.

398.—THE CIGAR PUZZLE.
I once propounded the following puzzle in a London club, and for a
considerable period it absorbed the attention of the members. They could
make nothing of it, and considered it quite impossible of solution. And yet,
as I shall show, the answer is remarkably simple.
Two men are seated at a square-topped table. One places an ordinary cigar
(flat at one end, pointed at the other) on the table, then the other does the
same, and so on alternately, a condition being that no cigar shall touch
another. Which player should succeed in placing the last cigar, assuming
that they each will play in the best possible manner? The size of the table
top and the size of the cigar are not given, but in order to exclude the
ridiculous answer that the table might be so diminutive as only to take one
cigar, we will say that the table must not be less than 2 feet square and the
cigar not more than 4½ inches long. With those restrictions you may take
any dimensions you like. Of course we assume that all the cigars are exactly
alike in every respect. Should the first player, or the second player, win?

MAGIC SQUARE PROBLEMS.
"By magic numbers."
CONGREVE, The Mourning Bride.
This is a very ancient branch of mathematical puzzledom, and it has an
immense, though scattered, literature of its own. In their simple form of
consecutive whole numbers arranged in a square so that every column,
every row, and each of the two long diagonals shall add up alike, these
magic squares offer three main lines of investigation: Construction,
Enumeration, and Classification. Of recent years many ingenious methods
have been devised for the construction of magics, and the law of their
formation is so well understood that all the ancient mystery has evaporated
and there is no longer any difficulty in making squares of any dimensions.
Almost the last word has been said on this subject. The question of the
enumeration of all the possible squares of a given order stands just where it
did over two hundred years ago. Everybody knows that there is only one
solution for the third order, three cells by three; and Frénicle published in
1693 diagrams of all the arrangements of the fourth order—880 in number
—and his results have been verified over and over again. I may here refer to
the general solution for this order, for numbers not necessarily consecutive,
by E. Bergholt in Nature, May 26, 1910, as it is of the greatest importance
to students of this subject. The enumeration of the examples of any higher
order is a completely unsolved problem.
As to classification, it is largely a matter of individual taste—perhaps an
æsthetic question, for there is beauty in the law and order of numbers. A
man once said that he divided the human race into two great classes: those
who take snuff and those who do not. I am not sure that some of our
classifications of magic squares are not almost as valueless. However,
lovers of these things seem somewhat agreed that Nasik magic squares (so
named by Mr. Frost, a student of them, after the town in India where he
lived, and also called Diabolique and Pandiagonal) and Associated magic
squares are of special interest, so I will just explain what these are for the
benefit of the novice.

I published in The Queen for January 15, 1910, an article that would enable
the reader to write out, if he so desired, all the 880 magics of the fourth
order, and the following is the complete classification that I gave. The first
example is that of a Simple square that fulfils the simple conditions and no
more. The second example is a Semi-Nasik, which has the additional
property that the opposite short diagonals of two cells each together sum to
34. Thus, 14 + 4 + 11 + 5 = 34 and 12 + 6 + 13 + 3 = 34. The third example
is not only Semi-Nasik but also Associated, because in it every number, if
added to the number that is equidistant, in a straight line, from the centre
gives 17. Thus, 1 + 16, 2 + 15, 3 + 14, etc. The fourth example, considered
the most "perfect" of all, is a Nasik. Here all the broken diagonals sum to
34. Thus, for example, 15 + 14 + 2 + 3, and 10 + 4 + 7 + 13, and 15 + 5 +
2 + 12. As a consequence, its properties are such that if you repeat the
square in all directions you may mark off a square, 4 × 4, wherever you
please, and it will be magic.
The following table not only gives a complete enumeration under the four
forms described, but also a classification under the twelve graphic types
indicated in the diagrams. The dots at the end of each line represent the
relative positions of those complementary pairs, 1 + 16, 2 + 15, etc., which
sum to 17. For example, it will be seen that the first and second magic
squares given are of Type VI., that the third square is of Type III., and that
the fourth is of Type I. Edouard Lucas indicated these types, but he dropped
exactly half of them and did not attempt the classification.

NASIK (Type I.) 48
SEMI-NASIK (Type II., Transpositions of Nasik) 48
" (Type III., Associated) 48
" (Type IV.) 96
" (Type V.) 96 192
" (Type VI.) 96 384
SIMPLE. (Type VI.) 208
" (Type VII.) 56
" (Type VIII.) 56
" (Type IX.) 56
" (Type X.) 56 224
" (Type XI.) 8
" (Type XII.) 8 16 448
880

It is hardly necessary to say that every one of these squares will produce
seven others by mere reversals and reflections, which we do not count as
different. So that there are 7,040 squares of this order, 880 of which are
fundamentally different.
An infinite variety of puzzles may be made introducing new conditions into
the magic square. In The Canterbury Puzzles I have given examples of such
squares with coins, with postage stamps, with cutting-out conditions, and
other tricks. I will now give a few variants involving further novel
conditions.

399.—THE TROUBLESOME EIGHT.
Nearly everybody knows that a "magic square" is an arrangement of
numbers in the form of a square so that every row, every column, and each
of the two long diagonals adds up alike. For example, you would find little
difficulty in merely placing a different number in each of the nine cells in
the illustration so that the rows, columns, and diagonals shall all add up 15.
And at your first attempt you will probably find that you have an 8 in one of
the corners. The puzzle is to construct the magic square, under the same
conditions, with the 8 in the position shown.

400.—THE MAGIC STRIPS.

I happened to have lying on my table a number of strips of cardboard, with
numbers printed on them from 1 upwards in numerical order. The idea
suddenly came to me, as ideas have a way of unexpectedly coming, to make
a little puzzle of this. I wonder whether many readers will arrive at the same
solution that I did.
Take seven strips of cardboard and lay them together as above. Then write
on each of them the numbers 1, 2, 3, 4, 5, 6, 7, as shown, so that the
numbers shall form seven rows and seven columns.
Now, the puzzle is to cut these strips into the fewest possible pieces so that
they may be placed together and form a magic square, the seven rows,
seven columns, and two diagonals adding up the same number. No figures
may be turned upside down or placed on their sides—that is, all the strips
must lie in their original direction.

Of course you could cut each strip into seven separate pieces, each piece
containing a number, and the puzzle would then be very easy, but I need
hardly say that forty-nine pieces is a long way from being the fewest
possible.

401.—EIGHT JOLLY GAOL BIRDS.

The illustration shows the plan of a prison of nine cells all communicating
with one another by doorways. The eight prisoners have their numbers on
their backs, and any one of them is allowed to exercise himself in
whichever cell may happen to be vacant, subject to the rule that at no time
shall two prisoners be in the same cell. The merry monarch in whose
dominions the prison was situated offered them special comforts one
Christmas Eve if, without breaking that rule, they could so place themselves
that their numbers should form a magic square.

Now, prisoner No. 7 happened to know a good deal about magic squares, so
he worked out a scheme and naturally selected the method that was most
expeditious—that is, one involving the fewest possible moves from cell to
cell. But one man was a surly, obstinate fellow (quite unfit for the society of
his jovial companions), and he refused to move out of his cell or take any
part in the proceedings. But No. 7 was quite equal to the emergency, and
found that he could still do what was required in the fewest possible moves
without troubling the brute to leave his cell. The puzzle is to show how he
did it and, incidentally, to discover which prisoner was so stupidly
obstinate. Can you find the fellow?

402.—NINE JOLLY GAOL BIRDS.

Shortly after the episode recorded in the last puzzle occurred, a ninth
prisoner was placed in the vacant cell, and the merry monarch then offered
them all complete liberty on the following strange conditions. They were
required so to rearrange themselves in the cells that their numbers formed a

magic square without their movements causing any two of them ever to be
in the same cell together, except that at the start one man was allowed to be
placed on the shoulders of another man, and thus add their numbers
together, and move as one man. For example, No. 8 might be placed on the
shoulders of No. 2, and then they would move about together as 10. The
reader should seek first to solve the puzzle in the fewest possible moves,
and then see that the man who is burdened has the least possible amount of
work to do.

403.—THE SPANISH DUNGEON.
Not fifty miles from Cadiz stood in the middle ages a castle, all traces of
which have for centuries disappeared. Among other interesting features, this
castle contained a particularly unpleasant dungeon divided into sixteen
cells, all communicating with one another, as shown in the illustration.
Now, the governor was a merry wight, and very fond of puzzles withal. One
day he went to the dungeon and said to the prisoners, "By my halidame!"
(or its equivalent in Spanish) "you shall all be set free if you can solve this
puzzle. You must so arrange yourselves in the sixteen cells that the numbers
on your backs shall form a magic square in which every column, every row,
and each of the two diagonals shall add up the same. Only remember this:
that in no case may two of you ever be together in the same cell."
One of the prisoners, after working at the problem for two or three days,
with a piece of chalk, undertook to obtain the liberty of himself and his
fellow-prisoners if they would follow his directions and move through the
doorway from cell to cell in the order in which he should call out their
numbers.

He succeeded in his attempt, and, what is more remarkable, it would seem
from the account of his method recorded in the ancient manuscript lying
before me, that he did so in the fewest possible moves. The reader is asked
to show what these moves were.

404.—THE SIBERIAN DUNGEONS.

The above is a trustworthy plan of a certain Russian prison in Siberia. All
the cells are numbered, and the prisoners are numbered the same as the cells
they occupy. The prison diet is so fattening that these political prisoners are
in perpetual fear lest, should their pardon arrive, they might not be able to
squeeze themselves through the narrow doorways and get out. And of
course it would be an unreasonable thing to ask any government to pull
down the walls of a prison just to liberate the prisoners, however innocent
they might be. Therefore these men take all the healthy exercise they can in
order to retard their increasing obesity, and one of their recreations will
serve to furnish us with the following puzzle.
Show, in the fewest possible moves, how the sixteen men may form
themselves into a magic square, so that the numbers on their backs shall add
up the same in each of the four columns, four rows, and two diagonals
without two prisoners having been at any time in the same cell together. I
had better say, for the information of those who have not yet been made
acquainted with these places, that it is a peculiarity of prisons that you are
not allowed to go outside their walls. Any prisoner may go any distance that
is possible in a single move.

405.—CARD MAGIC SQUARES.

Take an ordinary pack of cards and throw out the twelve court cards. Now,
with nine of the remainder (different suits are of no consequence) form the
above magic square. It will be seen that the pips add up fifteen in every row
in every column, and in each of the two long diagonals. The puzzle is with
the remaining cards (without disturbing this arrangement) to form three
more such magic squares, so that each of the four shall add up to a different
sum. There will, of course, be four cards in the reduced pack that will not be
used. These four may be any that you choose. It is not a difficult puzzle, but
requires just a little thought.

406.—THE EIGHTEEN DOMINOES.
The illustration shows eighteen dominoes arranged in the form of a square
so that the pips in every one of the six columns, six rows, and two long
diagonals add up 13. This is the smallest summation possible with any
selection of dominoes from an ordinary box of twenty-eight. The greatest
possible summation is 23, and a solution for this number may be easily
obtained by substituting for every number its complement to 6. Thus for
every blank substitute a 6, for every 1 a 5, for every 2 a 4, for 3 a 3, for 4 a
2, for 5 a 1, and for 6 a blank. But the puzzle is to make a selection of
eighteen dominoes and arrange them (in exactly the form shown) so that the
summations shall be 18 in all the fourteen directions mentioned.

SUBTRACTING, MULTIPLYING, AND
DIVIDING MAGICS.
Although the adding magic square is of such great antiquity, curiously
enough the multiplying magic does not appear to have been mentioned until
the end of the eighteenth century, when it was referred to slightly by one
writer and then forgotten until I revived it in Tit-Bits in 1897. The dividing
magic was apparently first discussed by me in The Weekly Dispatch in June
1898. The subtracting magic is here introduced for the first time. It will now
be convenient to deal with all four kinds of magic squares together.

In these four diagrams we have examples in the third order of adding,
subtracting, multiplying, and dividing squares. In the first the constant, 15,
is obtained by the addition of the rows, columns, and two diagonals. In the
second case you get the constant, 5, by subtracting the first number in a line
from the second, and the result from the third. You can, of course, perform
the operation in either direction; but, in order to avoid negative numbers, it
is more convenient simply to deduct the middle number from the sum of the
two extreme numbers. This is, in effect, the same thing. It will be seen that
the constant of the adding square is n times that of the subtracting square
derived from it, where n is the number of cells in the side of square. And the
manner of derivation here is simply to reverse the two diagonals. Both
squares are "associated"—a term I have explained in the introductory article
to this department.
The third square is a multiplying magic. The constant, 216, is obtained by
multiplying together the three numbers in any line. It is "associated" by

multiplication, instead of by addition. It is here necessary to remark that in
an adding square it is not essential that the nine numbers should be
consecutive. Write down any nine numbers in this way—

13 5
46 8
79 11

so that the horizontal differences are all alike and the vertical differences
also alike (here 2 and 3), and these numbers will form an adding magic
square. By making the differences 1 and 3 we, of course, get consecutive
numbers—a particular case, and nothing more. Now, in the case of the
multiplying square we must take these numbers in geometrical instead of
arithmetical progression, thus—

1 3 9
2 6 18
4 12 36

Here each successive number in the rows is multiplied by 3, and in the
columns by 2. Had we multiplied by 2 and 8 we should get the regular
geometrical progression, 1, 2, 4, 8, 16, 32, 64, 128, and 256, but I wish to
avoid high numbers. The numbers are arranged in the square in the same
order as in the adding square.
The fourth diagram is a dividing magic square. The constant 6 is here
obtained by dividing the second number in a line by the first (in either
direction) and the third number by the quotient. But, again, the process is
simplified by dividing the product of the two extreme numbers by the
middle number. This square is also "associated" by multiplication. It is
derived from the multiplying square by merely reversing the diagonals, and
the constant of the multiplying square is the cube of that of the dividing
square derived from it.
The next set of diagrams shows the solutions for the fifth order of square.
They are all "associated" in the same way as before. The subtracting square
is derived from the adding square by reversing the diagonals and
exchanging opposite numbers in the centres of the borders, and the constant

of one is again n times that of the other. The dividing square is derived from
the multiplying square in the same way, and the constant of the latter is the
5th power (that is the nth) of that of the former.

These squares are thus quite easy for odd orders. But the reader will
probably find some difficulty over the even orders, concerning which I will
leave him to make his own researches, merely propounding two little
problems.

407.—TWO NEW MAGIC SQUARES.
Construct a subtracting magic square with the first sixteen whole numbers
that shall be "associated" by subtraction. The constant is, of course,
obtained by subtracting the first number from the second in line, the result
from the third, and the result again from the fourth. Also construct a
dividing magic square of the same order that shall be "associated" by
division. The constant is obtained by dividing the second number in a line
by the first, the third by the quotient, and the fourth by the next quotient.

408.—MAGIC SQUARES OF TWO DEGREES.
While reading a French mathematical work I happened to come across, the
following statement: "A very remarkable magic square of 8, in two degrees,
has been constructed by M. Pfeffermann. In other words, he has managed to
dispose the sixty-four first numbers on the squares of a chessboard in such a
way that the sum of the numbers in every line, every column, and in each of
the two diagonals, shall be the same; and more, that if one substitutes for all
the numbers their squares, the square still remains magic." I at once set to
work to solve this problem, and, although it proved a very hard nut, one was
rewarded by the discovery of some curious and beautiful laws that govern
it. The reader may like to try his hand at the puzzle.

MAGIC SQUARES OF PRIMES.
The problem of constructing magic squares with prime numbers only was
first discussed by myself in The Weekly Dispatch for 22nd July and 5th
August 1900; but during the last three or four years it has received great
attention from American mathematicians. First, they have sought to form
these squares with the lowest possible constants. Thus, the first nine prime
numbers, 1 to 23 inclusive, sum to 99, which (being divisible by 3) is
theoretically a suitable series; yet it has been demonstrated that the lowest
possible constant is 111, and the required series as follows: 1, 7, 13, 31, 37,
43, 61, 67, and 73. Similarly, in the case of the fourth order, the lowest
series of primes that are "theoretically suitable" will not serve. But in every
other order, up to the 12th inclusive, magic squares have been constructed
with the lowest series of primes theoretically possible. And the 12th is the
lowest order in which a straight series of prime numbers, unbroken, from 1
upwards has been made to work. In other words, the first 144 odd prime
numbers have actually been arranged in magic form. The following
summary is taken from The Monist (Chicago) for October 1913:—

Order Totals Lowest Squares
of Square. of Series. Constants. made by—
Henry E.
3rd 333 111
Dudeney (1900).
Ernest Bergholt
4th 408 102
and C. D. Shuldham.
5th 1065 213 H. A. Sayles.
C. D. Shuldham
6th 2448 408
and J. N. Muncey.
7th 4893 699 do.
8th 8912 1114 do.
9th 15129 1681 do.
10th 24160 2416 J. N. Muncey.
11th 36095 3355 do.

12th 54168 4514 do.

For further details the reader should consult the article itself, by W. S.
Andrews and H. A. Sayles.
These same investigators have also performed notable feats in constructing
associated and bordered prime magics, and Mr. Shuldham has sent me a
remarkable paper in which he gives examples of Nasik squares constructed
with primes for all orders from the 4th to the 10th, with the exception of the
3rd (which is clearly impossible) and the 9th, which, up to the time of
writing, has baffled all attempts.

409.—THE BASKETS OF PLUMS.

This is the form in which I first introduced the question of magic squares
with prime numbers. I will here warn the reader that there is a little trap.
A fruit merchant had nine baskets. Every basket contained plums (all sound
and ripe), and the number in every basket was different. When placed as
shown in the illustration they formed a magic square, so that if he took any
three baskets in a line in the eight possible directions there would always be
the same number of plums. This part of the puzzle is easy enough to
understand. But what follows seems at first sight a little queer.
The merchant told one of his men to distribute the contents of any basket he
chose among some children, giving plums to every child so that each should
receive an equal number. But the man found it quite impossible, no matter
which basket he selected and no matter how many children he included in
the treat. Show, by giving contents of the nine baskets, how this could come
about.

410.—THE MANDARIN'S "T" PUZZLE.

Before Mr. Beauchamp Cholmondely Marjoribanks set out on his tour in
the Far East, he prided himself on his knowledge of magic squares, a
subject that he had made his special hobby; but he soon discovered that he
had never really touched more than the fringe of the subject, and that the
wily Chinee could beat him easily. I present a little problem that one
learned mandarin propounded to our traveller, as depicted on the last page.
The Chinaman, after remarking that the construction of the ordinary magic
square of twenty-five cells is "too velly muchee easy," asked our
countryman so to place the numbers 1 to 25 in the square that every
column, every row, and each of the two diagonals should add up 65, with
only prime numbers on the shaded "T." Of course the prime numbers
available are 1, 2, 3, 5, 7, 11, 13, 17, 19, and 23, so you are at liberty to
select any nine of these that will serve your purpose. Can you construct this
curious little magic square?

411.—A MAGIC SQUARE OF COMPOSITES.
As we have just discussed the construction of magic squares with prime
numbers, the following forms an interesting companion problem. Make a
magic square with nine consecutive composite numbers—the smallest
possible.

412.—THE MAGIC KNIGHT'S TOUR.
Here is a problem that has never yet been solved, nor has its impossibility
been demonstrated. Play the knight once to every square of the chessboard
in a complete tour, numbering the squares in the order visited, so that when
completed the square shall be "magic," adding up to 260 in every column,
every row, and each of the two long diagonals. I shall give the best answer
that I have been able to obtain, in which there is a slight error in the
diagonals alone. Can a perfect solution be found? I am convinced that it
cannot, but it is only a "pious opinion."

MAZES AND HOW TO THREAD THEM.
"In wandering mazes lost."
Paradise Lost.
The Old English word "maze," signifying a labyrinth, probably comes from
the Scandinavian, but its origin is somewhat uncertain. The late Professor
Skeat thought that the substantive was derived from the verb, and as in old
times to be mazed or amazed was to be "lost in thought," the transition to a
maze in whose tortuous windings we are lost is natural and easy.
The word "labyrinth" is derived from a Greek word signifying the passages
of a mine. The ancient mines of Greece and elsewhere inspired fear and awe
on account of their darkness and the danger of getting lost in their intricate
passages. Legend was afterwards built round these mazes. The most
familiar instance is the labyrinth made by Dædalus in Crete for King Minos.
In the centre was placed the Minotaur, and no one who entered could find
his way out again, but became the prey of the monster. Seven youths and
seven maidens were sent regularly by the Athenians, and were duly
devoured, until Theseus slew the monster and escaped from the maze by aid
of the clue of thread provided by Ariadne; which accounts for our using to-
day the expression "threading a maze."
The various forms of construction of mazes include complicated ranges of
caverns, architectural labyrinths, or sepulchral buildings, tortuous devices
indicated by coloured marbles and tiled pavements, winding paths cut in the
turf, and topiary mazes formed by clipped hedges. As a matter of fact, they
may be said to have descended to us in precisely this order of variety.
Mazes were used as ornaments on the state robes of Christian emperors
before the ninth century, and were soon adopted in the decoration of
cathedrals and other churches. The original idea was doubtless to employ
them as symbols of the complicated folds of sin by which man is
surrounded. They began to abound in the early part of the twelfth century,
and I give an illustration of one of this period in the parish church at St.
Quentin (Fig. 1). It formed a pavement of the nave, and its diameter is 34½

feet. The path here is the line itself. If you place your pencil at the point A
and ignore the enclosing line, the line leads you to the centre by a long
route over the entire area; but you never have any option as to direction
during your course. As we shall find in similar cases, these early
ecclesiastical mazes were generally not of a puzzle nature, but simply long,
winding paths that took you over practically all the ground enclosed.

FIG. 1.—Maze at St. Quentin.
In the abbey church of St. Berlin, at St. Omer, is another of these curious
floors, representing the Temple of Jerusalem, with stations for pilgrims.
These mazes were actually visited and traversed by them as a compromise
for not going to the Holy Land in fulfilment of a vow. They were also used
as a means of penance, the penitent frequently being directed to go the
whole course of the maze on hands and knees.

FIG. 2.—Maze in Chartres Cathedral.
The maze in Chartres Cathedral, of which I give an illustration (Fig. 2), is
40 feet across, and was used by penitents following the procession of
Calvary. A labyrinth in Amiens Cathedral was octagonal, similar to that at
St. Quentin, measuring 42 feet across. It bore the date 1288, but was
destroyed in 1708. In the chapter-house at Bayeux is a labyrinth formed of
tiles, red, black, and encaustic, with a pattern of brown and yellow. Dr.
Ducarel, in his "Tour through Part of Normandy" (printed in 1767),
mentions the floor of the great guard-chamber in the abbey of St. Stephen,
at Caen, "the middle whereof represents a maze or labyrinth about 10 feet
diameter, and so artfully contrived that, were we to suppose a man
following all the intricate meanders of its volutes, he could not travel less
than a mile before he got from one end to the other."

FIG. 3.—Maze in Lucca Cathedral.
Then these mazes were sometimes reduced in size and represented on a
single tile (Fig. 3). I give an example from Lucca Cathedral. It is on one of
the porch piers, and is 19½ inches in diameter. A writer in 1858 says that,
"from the continual attrition it has received from thousands of tracing
fingers, a central group of Theseus and the Minotaur has now been very
nearly effaced." Other examples were, and perhaps still are, to be found in
the Abbey of Toussarts, at Châlons-sur-Marne, in the very ancient church of
St. Michele at Pavia, at Aix in Provence, in the cathedrals of Poitiers,
Rheims, and Arras, in the church of Santa Maria in Aquiro in Rome, in San
Vitale at Ravenna, in the Roman mosaic pavement found at Salzburg, and
elsewhere. These mazes were sometimes called "Chemins de Jerusalem," as
being emblematical of the difficulties attending a journey to the earthly
Jerusalem and of those encountered by the Christian before he can reach the
heavenly Jerusalem—where the centre was frequently called "Ciel."
Common as these mazes were upon the Continent, it is probable that no
example is to be found in any English church; at least I am not aware of the
existence of any. But almost every county has, or has had, its specimens of

mazes cut in the turf. Though these are frequently known as "miz-mazes" or
"mize-mazes," it is not uncommon to find them locally called "Troy-towns,"
"shepherds' races," or "Julian's Bowers"—names that are misleading, as
suggesting a false origin. From the facts alone that many of these English
turf mazes are clearly copied from those in the Continental churches, and
practically all are found close to some ecclesiastical building or near the site
of an ancient one, we may regard it as certain that they were of church
origin and not invented by the shepherds or other rustics. And curiously
enough, these turf mazes are apparently unknown on the Continent. They
are distinctly mentioned by Shakespeare:—

"The nine men's morris is filled up with mud,
And the quaint mazes in the wanton green
For lack of tread are undistinguishable."

A Midsummer Night's Dream, ii. 1.

"My old bones ache: here's a maze trod indeed,
Through forth-rights and meanders!"

The Tempest, iii. 3.

There was such a maze at Comberton, in Cambridgeshire, and another,
locally called the "miz-maze," at Leigh, in Dorset. The latter was on the
highest part of a field on the top of a hill, a quarter of a mile from the
village, and was slightly hollow in the middle and enclosed by a bank about
3 feet high. It was circular, and was thirty paces in diameter. In 1868 the
turf had grown over the little trenches, and it was then impossible to trace
the paths of the maze. The Comberton one was at the same date believed to
be perfect, but whether either or both have now disappeared I cannot say.
Nor have I been able to verify the existence or non-existence of the other
examples of which I am able to give illustrations. I shall therefore write of
them all in the past tense, retaining the hope that some are still preserved.

FIG. 4.—Maze at Saffron Walden, Essex.
In the next two mazes given—that at Saffron Walden, Essex (110 feet in
diameter, Fig. 4), and the one near St. Anne's Well, at Sneinton,
Nottinghamshire (Fig. 5), which was ploughed up on February 27th, 1797
(51 feet in diameter, with a path 535 yards long)—the paths must in each
case be understood to be on the lines, black or white, as the case may be.

FIG. 5.—Maze at Sneinton, Nottinghamshire.

FIG. 6.—Maze at Alkborough, Lincolnshire.

I give in Fig. 6 a maze that was at Alkborough, Lincolnshire, overlooking
the Humber. This was 44 feet in diameter, and the resemblance between it
and the mazes at Chartres and Lucca (Figs. 2 and 3) will be at once
perceived. A maze at Boughton Green, in Nottinghamshire, a place
celebrated at one time for its fair (Fig. 7), was 37 feet in diameter. I also
include the plan (Fig. 8) of one that used to be on the outskirts of the village
of Wing, near Uppingham, Rutlandshire. This maze was 40 feet in diameter.

FIG. 7.—Maze at Boughton Green,
Nottinghamshire.

FIG. 8.—Maze at Wing, Rutlandshire.

FIG. 9.—Maze on St. Catherine's Hill,
Winchester.
The maze that was on St. Catherine's Hill, Winchester, in the parish of
Chilcombe, was a poor specimen (Fig. 9), since, as will be seen, there was
one short direct route to the centre, unless, as in Fig. 10 again, the path is
the line itself from end to end. This maze was 86 feet square, cut in the turf,
and was locally known as the "Mize-maze." It became very indistinct about
1858, and was then recut by the Warden of Winchester, with the aid of a
plan possessed by a lady living in the neighbourhood.

FIG. 10.—Maze on Ripon Common.
A maze formerly existed on Ripon Common, in Yorkshire (Fig. 10). It was
ploughed up in 1827, but its plan was fortunately preserved. This example
was 20 yards in diameter, and its path is said to have been 407 yards long.

FIG. 11.—Maze at Theobalds, Hertfordshire.
In the case of the maze at Theobalds, Hertfordshire, after you have found
the entrance within the four enclosing hedges, the path is forced (Fig. 11).
As further illustrations of this class of maze, I give one taken from an
Italian work on architecture by Serlio, published in 1537 (Fig. 12), and one
by London and Wise, the designers of the Hampton Court maze, from their
book, The Retired Gard'ner, published in 1706 (Fig. 13). Also, I add a
Dutch maze (Fig. 14).

FIG. 12.—Italian Maze of Sixteenth Century.

FIG. 13.—By the Designers of Hampton Court
Maze.

FIG. 14.—A Dutch Maze.
So far our mazes have been of historical interest, but they have presented no
difficulty in threading. After the Reformation period we find mazes
converted into mediums for recreation, and they generally consisted of
labyrinthine paths enclosed by thick and carefully trimmed hedges. These
topiary hedges were known to the Romans, with whom the topiarius was
the ornamental gardener. This type of maze has of late years degenerated
into the seaside "Puzzle Gardens. Teas, sixpence, including admission to the
Maze." The Hampton Court Maze, sometimes called the "Wilderness," at
the royal palace, was designed, as I have said, by London and Wise for
William III., who had a liking for such things (Fig. 15). I have before me
some three or four versions of it, all slightly different from one another; but
the plan I select is taken from an old guide-book to the palace, and therefore
ought to be trustworthy. The meaning of the dotted lines, etc., will be
explained later on.

FIG. 15.—Maze at Hampton Court Palace.

FIG. 16.—Maze at Hatfield House, Herts.
The maze at Hatfield House (Fig. 16), the seat of the Marquis of Salisbury,
like so many labyrinths, is not difficult on paper; but both this and the
Hampton Court Maze may prove very puzzling to actually thread without
knowing the plan. One reason is that one is so apt to go down the same
blind alleys over and over again, if one proceeds without method. The maze
planned by the desire of the Prince Consort for the Royal Horticultural
Society's Gardens at South Kensington was allowed to go to ruin, and was

then destroyed—no great loss, for it was a feeble thing. It will be seen that
there were three entrances from the outside (Fig. 17), but the way to the
centre is very easy to discover. I include a German maze that is curious, but
not difficult to thread on paper (Fig. 18). The example of a labyrinth
formerly existing at Pimperne, in Dorset, is in a class by itself (Fig. 19). It
was formed of small ridges about a foot high, and covered nearly an acre of
ground; but it was, unfortunately, ploughed up in 1730.

FIG. 17.—Maze formerly at South
Kensington.

FIG. 18.—A German Maze.

FIG. 19.—Maze at Pimperne, Dorset.
We will now pass to the interesting subject of how to thread any maze.
While being necessarily brief, I will try to make the matter clear to readers
who have no knowledge of mathematics. And first of all we will assume
that we are trying to enter a maze (that is, get to the "centre") of which we

have no plan and about which we know nothing. The first rule is this: If a
maze has no parts of its hedges detached from the rest, then if we always
keep in touch with the hedge with the right hand (or always touch it with
the left), going down to the stop in every blind alley and coming back on
the other side, we shall pass through every part of the maze and make our
exit where we went in. Therefore we must at one time or another enter the
centre, and every alley will be traversed twice.
Now look at the Hampton Court plan. Follow, say to the right, the path
indicated by the dotted line, and what I have said is clearly correct if we
obliterate the two detached parts, or "islands," situated on each side of the
star. But as these islands are there, you cannot by this method traverse every
part of the maze; and if it had been so planned that the "centre" was, like the
star, between the two islands, you would never pass through the "centre" at
all. A glance at the Hatfield maze will show that there are three of these
detached hedges or islands at the centre, so this method will never take you
to the "centre" of that one. But the rule will at least always bring you safely
out again unless you blunder in the following way. Suppose, when you were
going in the direction of the arrow in the Hampton Court Maze, that you
could not distinctly see the turning at the bottom, that you imagined you
were in a blind alley and, to save time, crossed at once to the opposite
hedge, then you would go round and round that U-shaped island with your
right hand still always on the hedge—for ever after!
This blunder happened to me a few years ago in a little maze on the isle of
Caldy, South Wales. I knew the maze was a small one, but after a very long
walk I was amazed to find that I did not either reach the "centre" or get out
again. So I threw a piece of paper on the ground, and soon came round to it;
from which I knew that I had blundered over a supposed blind alley and
was going round and round an island. Crossing to the opposite hedge and
using more care, I was quickly at the centre and out again. Now, if I had
made a similar mistake at Hampton Court, and discovered the error when at
the star, I should merely have passed from one island to another! And if I
had again discovered that I was on a detached part, I might with ill luck
have recrossed to the first island again! We thus see that this "touching the
hedge" method should always bring us safely out of a maze that we have
entered; it may happen to take us through the "centre," and if we miss the
centre we shall know there must be islands. But it has to be done with a