Amusements in Mathematics by Henry Ernest Dudeney

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Title: Amusements in Mathematics

Author: Henry Ernest Dudeney

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Language: English

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[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.

Exponents are represented in this text by ^, e.g. '3 squared' is 3^2.

Numbers with fractional components (other than ¼, ½ and ¾) have a +
symbol separating the whole number component from the fraction. It makes
the fraction look odd, but yeilds correct solutions no matter how it is
interpreted. E.G., 4 and eleven twenty-thirds is 4+11/23, not 411/23 or
4-11/23.

]




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 _Hudibras_.


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 105
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.

[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.--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.

[Illustration]

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?

[Illustration]


63.--THE STOP-WATCH.

[Illustration]

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?"

[Illustration]


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.

[Illustration:

( 15 Gals )

(31 Gals) (19 Gals)

(20 Gals) (16 Gals) (18 Gals)

]


77.--DIGITS AND SQUARES.

[Illustration:

+---+---+---+
| 1 | 9 | 2 |
+---+---+---+
| 3 | 8 | 4 |
+---+---+---+
| 5 | 7 | 6 |
+---+---+---+

]

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.

[Illustration:

A B C
================== ================== ==================
| +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ |
| | | | | | | | | | | | | | | | | | | | | | | |
| +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ |
| | | | | |
| +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ |
| | | | | | | | | | | | | | | | | | | | | | | |
| +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ |
| | | | | |
================== ================== ==================
| +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ |
| | | | | | | | | | | | | | | | | | | | | | | |
| +--+ +--+ +--+ | | +--+ +--+ +--+ | | +--+ +--+ +--+ |
------------------ ------------------ ------------------

]

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.

[Illustration:

(1)(5)(8) (7)(9)
(2)(3) (4)(6)

]

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, 7 × 658 and 14 × 329 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.

[Illustration]

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.

[Illustration]

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 0 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.

[Illustration]

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 99+9/9, but the four 7's call for rather more
ingenuity. Can you discover the little trick?


96.--THE DICE NUMBERS.

[Illustration]

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.

[Illustration]

"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.

[Illustration]

"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.

[Illustration]

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.

[Illustration]

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 0 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 5^4×2^3, 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 5^4×2^3
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.

[Illustration]

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.

[Illustration]

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.

[Illustration]

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.

[Illustration]


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.

[Illustration]

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.

[Illustration: [Pyramid of cannon-balls]]


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"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?

[Illustration]


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.

[Illustration: Fig. 1]

[Illustration: Fig. 2]

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.

[Illustration: Fig. 3]

[Illustration: Fig. 4]

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.

[Illustration: Fig. 5.]

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.

[Illustration: Fig. 6]

[Illustration: Fig. 7]

[Illustration: Fig. 8]

[Illustration: Fig. 9]

[Illustration: Fig. 10]

[Illustration: Fig. 11]

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.

[Illustration: FIG. 12.]

[Illustration: FIG. 13.]

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.

[Illustration: FIG. 14.]

[Illustration: FIG. 15.]

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.

[Illustration: FIG. 16.]

[Illustration: FIG. 17.]

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.

[Illustration: FIG. 18.]

[Illustration: FIG. 19.]

[Illustration: FIG. 20.]

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.

[Illustration: FIG. 21.]

[Illustration: FIG. 22.]

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.

[Illustration: FIG. 23.]

[Illustration: FIG. 24.]

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.

[Illustration: FIG. 25.]

[Illustration: FIG. 26.]

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.

[Illustration: FIG. 27.]

[Illustration: FIG. 28.]

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.

[Illustration: FIG. 29.]

[Illustration: FIG. 30.]

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.

[Illustration: FIG. 31.]

[Illustration: FIG. 32.]

[Illustration: FIG. 33.]

[Illustration: FIG. 34.]

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.

[Illustration: FIG. 35.]

[Illustration: FIG. 36.]

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.

[Illustration: FIG. 37.]

[Illustration: FIG. 38.]

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.

[Illustration: FIG. 39]

[Illustration: 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.

[Illustration: FIG. 41]

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.

[Illustration]

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 61+5/7
(instead of 63), then the step method would apply. For the steps would
then be 10+2/7 in. in height and 12 in. in breadth. Note that 61+5/7 ×
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.

[Illustration]

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.

[Illustration]

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.

[Illustration]


154.--MRS. HOBSON'S HEARTHRUG.

[Illustration]

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.

[Illustration]

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

[Illustration]


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.

[Illustration]

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.

[Illustration]

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.

[Illustration]


160.--THE TWO HORSESHOES.

[Illustration]

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.

[Illustration]

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.

[Illustration:

·-----------·-----------·-----------·-----------·
| · · ·|||||||· ·|||||||· · · |
| · ·|||· ·|||· · · |
| · · · · |
| · · · · · · · · |
| · · · · · · · · |
· · · · ·
||· · · · · /|\ · ·||
||||· · · · · \|/ · ·||||
||||||· · · ·||||||
||||· · · · · · · ·||||
||· · · · · · · ·||
· · · · ·
||· · · · · · · ·||
||||· · · · · · · ·||||
||||||· · · ·||||||
||||· · · · · · · ·||||
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· · · · ·
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| · ·|||· ·|||· · · |
| · · ·|||||||· ·|||||||· · · |
·-----------·-----------·-----------·-----------·

]

[Illustration]


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?

[Illustration:

--------
/ \ 1/ \
/ \ 2 \/ 3 / \
/ \ /\ / \
/ \ 4 \/ 5\/ 6 / \
| \ /\ /\ / |
\ 7\/ 8\/ 9\/10 /
\ /\ /\ /\ /
\/11\/12\/13\/
\ /\ /\ /
\/14\/15\/
\ /\ /
\/16\/
-----

]

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


165.--THE SEVEN PIGS.

[Illustration]

+------------------------------+
| |
| P |
| |
| P |
| P |
| P |
| P |
| P |
| P |
| |
+------------------------------+

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.

[Illustration]

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.

[Illustration]

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.

[Illustration]

"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.

[Illustration]

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.

[Illustration: 2]

[Illustration: 3]

[Illustration: 4]

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."

[Illustration: 5]

"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."

[Illustration: 6]

[Illustration: 7]

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 2+1/8 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.

[Illustration: 8]

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.

[Illustration: 9]

[Illustration: 10]

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.

[Illustration: 11]

[Illustration: 12]

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?

[Illustration: 13]

[Illustration: 14]

[Illustration: 15]

[Illustration: 16]



PATCHWORK PUZZLES.

"Of shreds and patches."--_Hamlet_, iii. 4.


170.--THE CUSHION COVERS.

[Illustration]

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.

[Illustration]

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.

[Illustration]


173.--MRS. PERKINS'S QUILT.

[Illustration]

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.

[Illustration]

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."

[Illustration]

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.

[Illustration]

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 12 × 12 and the
other 5 × 5. She proposes to join them together and make one square
patchwork quilt, 13 × 13, 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.

[Illustration]

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.

[Illustration]

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.

[Illustration]


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?

[Illustration]


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?

[Illustration]


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.

[Illustration]

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.

[Illustration]

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.

[Illustration]

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.

[Illustration]

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.

[Illustration]

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.

[Illustration]

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 SHEEPFOLD.

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.

[Illustration]

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.

[Illustration]


196.--THE TETHERED GOAT.

[Illustration]

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?

[Illustration]


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.

[Illustration]

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.

[Illustration]

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.

[Illustration]

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.

[Illustration]

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.

[Illustration]

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.

[Illustration]


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.

[Illustration]


207.--CHERRIES AND PLUMS.

[Illustration]

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.

[Illustration]

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.

[Illustration]


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?

[Illustration]


212.--THE BURMESE PLANTATION.

[Illustration]

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."

[Illustration]


214.--THE SIX FROGS.

[Illustration]

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?

[Illustration]


216.--THE EDUCATED FROGS.

[Illustration]

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
successive 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.

[Illustration:


( I ) ((N))

( M ) ((A))

( H ) ((T))

( E ) ((W))

( C ) ((K))
( )


]

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