Part II

Contents

PAGE
Optical Illusions
Freaks of Figures
Chess Cameos
Science at Play
Curious Calculations
Word and Letter Puzzles
Solutions

Optical Illusions

No. I.—Swallowed!

Take a small card and place it on its longer edge upon the dotted line. Now set the picture in a good light on the table, and let your head drop gradually towards the card until you almost touch it with your nose. You will see the bird fly into the jaws of the snake!

Amusing Problems

The Carpenter’s Puzzle

1. A carpenter was called in to mend a hole in a wooden floor. The gap was two feet wide, and twelve feet long, while the only board at hand was three feet wide, and eight feet long.

This had been put aside as useless, but, on catching sight of it, the carpenter ran his rule over it and said that he could make a perfect fit, and cover all the hole by cutting the board into two pieces. How did he do this?

Solution

No. Ii.—An Illusion of Rotation

This most interesting optical illusion was devised by Professor Thompson some years ago:—

If the illustration is moved by hand in a small circle on the level, with such motion as is given in rinsing out a bowl, the circles of the larger diagram will seem to revolve in the direction in which the paper is moved, while the cogs of the smaller diagram will apparently turn slowly in the opposite direction.

No. Iii.—Whirling Wheels

Here is another combination of the clever illusion of the whirling wheels.

If a rapid rotating motion is given to the diagram, each circle will seem to revolve, and the cog wheel in the centre will appear to move slowly round in the opposite direction.

Golden Pippins

2. A man leaves an orchard of forty choice apple trees to his ten sons. On the first tree is one apple, on the second there are two, on the third three, and so on to the fortieth, on which there are forty.

Each son is to have four of the trees, and on them an equal number of the apples. How can they thus apportion the trees, and how many apples will each son have? Here is one way:—

1	2	3	4	5	6	7	8	9	10
20	19	18	17	16	15	14	13	12	11
21	22	23	24	25	26	27	28	29	30
40	39	38	37	36	35	34	33	32	31
82	82	82	82	82	82	82	82	82	82

Can you find another perfect solution?

Solution

No. Iv.—An Illusion of Motion

We call very particular attention to this fascinating illustration of the fact that the mind and eye may receive and register false impressions under quite simple conditions:—

Hold this at rather more than reading distance, upright, and move it steadily up and down. The dark line will soon seem to slide up and down upon the perpendicular line. It will be better seen if drawn to pattern on a card.

No. V.—Are They Parallel?

As the eye falls upon the principal lines of this interesting diagram, an immediate impression is formed that they are not parallel.

This, however, is a most curious illusion, created in the mind entirely by the short sloping lines, as is found at once by the simple test of measurement.

An Awkward Fix

3. With no knowledge of the surrounding district, I was making my way to a distant town through country roads, guided by the successive sign-posts that were provided.

Coming presently to four cross-roads I found to my dismay that some one had in mischief uprooted the sign-post and thrown it into the ditch. In this perplexing fix how could I find my way? A bright thought struck me. What was it?

Solution

A Terrible Tumble

9. From what height must a man fall out of an airship—screaming as he goes overboard—so as to reach the earth before the sound of his cry?

N.B.—Resistance of the air, and the acoustical fact that sound will not travel from a rare to a dense atmosphere, are to be disregarded.

Solution

In a Predicament

10. Imagine a man on a perfectly smooth table surface of considerable size, in a vacuum, where there is no outside force to move him, and there is no friction. He may raise himself up and down, slide his feet about, double himself up, wave his arms, but his centre of gravity will be always vertically above the same point of the surface.

No. Xii.—The Fickle Barrel

Here is another excellent optical illusion. Look attentively at the diagram below, and notice in which direction you apparently look into it, as though it were an open cask.

Now shake the paper, or move it slightly, and you will find, more often than not, that you seem to see into it in quite the opposite direction.

Heads I Win!

15. I hold a penny level between my finger and thumb, and presently let it fall from the thumb by withdrawing my finger. It makes exactly a half-turn in falling through the first foot. If it starts “heads,” how far must it fall to bring it “heads” to the floor?

Solution

He Did It!

16. “They call these safety matches,” said Funnyboy at his club one day, “and say that they strike only on the box. Don’t believe it! I can strike them quite easily on my boot.”

No sooner said than done. He took out a match, struck it on his boot, and—phiz!—it was instantly alight. The box was handed round, and match after match was struck by the bystanders on their boots, but not one of them could succeed.

“You don’t give the magic touch,” said Funnyboy, as he gaily struck another. How did he do it?

Solution

No. Xiii.—A Strange Optical Illusion

How many cubes can you see as you look at the large diagram? The two smaller ones should be looked at first alternately, and they will assist the eye to see at one time six, and at another time seven, very distinct cubes.

No. Xiv.—A Circular Illusion

This curious optical illusion is not easily followed by eye to the finish of the several lines.

Each short line is, in fact, part of the circumference of a circle, and the circles when completed will be found to be accurately concentric. It would seem at first sight that the lines are taking courses which would eventually meet at some point common to them all.

A Cycle Surprise

17. We commend this curious point to the special attention of cyclists:—

A bicycle is stationary, with one pedal at its lowest point. If this bicycle is lightly supported, and the bottom pedal is pulled backward, what will happen?

Solution

A Handy Short Cut

Here is a delightfully simple way in which market gardeners, or others who buy or sell weighty produce, can check their invoices for potatoes or what not.

Say, for example, that a consignment weighs 6 tons, 10 cwts., 1 qr. Then, since 20 cwts. are to a ton as 20s. are to a pound, and each quarter would answer on these lines to 3d., we can at once write down £6 10s. 3d., as the price at £1 the ton. On this sure basis any further calculation is easily made.

No. Xxi.—The Persistence of Vision

HOW TO SEE THE GHOST

Look steadily, in a good light, for thirty seconds at the cross in the eye of the pictured skull; then look up at the wall or ceiling, or look fixedly at a sheet of paper for another thirty seconds, when a ghost-like image of the skull will be developed.

Not So Fast!

A gardener, when he had planted 100 trees on a line at intervals of 10 yards, was able to walk from the first of these to the last in a few seconds, for they were set on the circumference of a circle!

No. Xxii.—The Persistence of Vision

Here is another example of what is known as the persistence of vision:—

Look fixedly for some little time at this grotesque figure, then turn your eyes to the wall or ceiling, and you will in a few seconds see it appear in dark form upon a light ground.

A Puzzle Number

The sum of nine figures a number will make,
Of which just a third will remain
If fifty away from the whole you should take,
Thus turning a loss to a gain.

It needs something more than mere arithmetic to discover that the solution to this puzzle is XLV, the sum of the nine digits, for if the L is removed, XV, the third of XLV, remains.

No. Xxiii.—Another Parallel Freak

Here is another curious illusion:—

The four straight lines are perfectly parallel, but the contradictory herring-bones disturb the eye.

The Magic of Figures

If our penny had been current coin in the first year of the Christian era, and had been invested at compound interest at five per cent., it would have amounted in 1905 to more than £132,010,000,000,000,000,000,000,000,000,000,000,000.

This gigantic sum would afford an income of £101,890,000,000,000,000,000 every second to every man, woman, and child in the world, if we take its population to be 1,483,000,000 souls!

Absurdly small in contrast to these startling figures is the modest eight shillings which the same penny would have yielded in the same time at simple interest.

No. Xxiv.—Illusion of Length

Here in another form is shown the illusion of length.

At first sight it seems that the two upright lines are distinctly longer than the line that slopes, but it is not so.

What Is Your Age?

Here is a neat method of discovering the age of a person older than yourself:—

Subtract your own age from 99. Ask your friend to add this remainder to his age, and then to remove the first figure and add it to the last, telling you the result. This will always be the difference of your ages. Thus, if you are 22, and he is 35, 99 - 22 = 77. Then 35 + 77 = 112. The next process turns this into 13, which, added to your age, gives his age, 35.

No. Xxv.—The Honeycomb Illusion

In this diagram one hundred and twenty-one circular spots are grouped in a diamond.

If we half close our eyes, and look at this through our eye-lashes, we find that it takes on the appearance of a section of honeycomb, with hexagonal cells.

Multiplication No Vexation

Here is a ready method for multiplying together any two numbers between 12 and 20.

Take one of the two numbers and add it to the unit digit of the other. Beneath the sum thus obtained, but one place to the right, put the product of the unit digits of the two original numbers.

The sum of these new numbers is the product of the numbers that were chosen. Thus:—

19 × 13.	(19 + 3)	=	22
(9 × 3)		=	27
			247

No. Xxvi.—Chess Cameo By Dr Gold A Chess Joke

WHITE

White to play, and mate in two moves.

There are twelve variations in this beautiful problem.

Solution

Magic Multiplication

It will interest all who study short cuts and contrivances to know that a novice at arithmetic who has mastered simple addition, and can multiply or divide by 2, but by no higher numbers, can, by using all these methods, multiply any two numbers together easily and accurately.

This is how it is done:—

Write down the numbers, say 53 and 21, divide one of them by 2 as often as possible, omitting remainders, and multiply the other by 2 the same number of times; set these down side by side, as in the instance given below, and wherever there is an even number on the division side, strike out the corresponding number on the multiplication side. Add up what remains on that side, and the sum is done. Thus:—

53	21)
26	(42)
13	84)
6	(168)
3	336)
1	672)
	1113)

which is 53 multiplied by 21.

No. Xxxi.—Chess Cameo By A. W. Galitzky A Prize Problem

BLACK

WHITE

White to play, and mate in three moves.

Solution

A Smart Schoolboy

The question, “How many times can 19 be subtracted from a million?” was set by an examiner, who no doubt expected that the answer would be obtained by dividing a million by 19. One bright youth, however, filled a neatly-written page with repetitions of

1,000,000	1,000,000	1,000,000
19	19	19
999,981	999,981	999,981

and added at the foot of the page, “N.B.—I can do this as often as you like.”

There was a touch of unintended humour in this, for, after all, the boy gave a correct answer to a badly worded question.

No. Xlvii.—Chess Cameo By A. Cyril Pearson

BLACK

WHITE

White to play, and mate in three moves.

Solution

No. Xlviii.—Chess Cameo By Frank Healey Quite a Gem

BLACK

WHITE

White to play, and mate in three moves.

Solution

Lewis Carroll’s Short Cut

Here is a very smart and very simple method of dividing any multiple of 9 by 9, from the fertile brain of Lewis Carroll:—Place a cypher over the final figure, subtract the final figure from this, place the result above in the tens place, subtract the original tens figure from this, and so on to the end. Then the top line, excluding the intruded cypher, gives the result desired. Thus:—

Here is a curious rough rule for remembering distances and sizes:—

The diameter of the earth multiplied by 108 gives approximately the sun’s diameter. The diameter of the sun multiplied by 108 gives the mean distance of the earth from the sun. The diameter of the moon multiplied by 108 gives the mean distance of the moon from the earth.

No. Lii.—Chess Cameo By J. Berger

BLACK

WHITE

White to play, and mate in three moves.

Solution

Personal Arithmetic

Says Giles, “My wife and I are two,
Yet faith I know not why, sir.”
Quoth Jack, “You’re ten, if I speak true,
She’s one, and you’re a cypher!”

No. Liii.—Chess Cameo By H. F. L. Meyer

BLACK

WHITE

White to play, and mate in three moves.

Solution

Division by Subtraction

Here is a curious and quite uncommon method of dividing any multiple of 11 by 11.

Set down the multiple of 11, place a cypher under its last figure, draw a line, and subtract, placing the first remainder under the tens place. Subtract this from the next number in order, and so on throughout, adding in always any number that is carried. Thus:—

363	56408	375034
0	0	0
33	5128	34094

No. Liv.—Chess Cameo By Frank Healey

BLACK

WHITE

White to play, and mate in three moves.

Solution

Luck in Odd Numbers

Perhaps the old saying, “there is luck in odd numbers,” may have some connection with the curious fact that the sum of any quantity of consecutive odd numbers, beginning always with 1, is the square of that number. Thus:—

1 + 3 + 5	=	9	=	3	×	3.
1 + 3 + 5, etc., up to 17	=	81	=	9	×	9.
1 + 3 + 5, etc., up to 99	=	2500	=	50	×	50.

No. Lv.—Chess Cameo By Frank Healey

BLACK

WHITE

White to play, and mate in three moves.

Solution

The Versatile Number

In the number 142857, if the digits which belong to it are in succession transposed from the first place to the end, the result is in each case a multiple of the original number. Thus:—

285714	=	142857	×	2
428571	=	142857	×	3
571428	=	142857	×	4
714285	=	142857	×	5
857142	=	142857	×	6

No. Lvi.—Chess Cameo By J. E. Campbell

BLACK

WHITE

White to play, and mate in three moves.

Solution

A Paradox

By the following simple method, a plausible attempt is made to prove that 1 is equal to 2:—

Suppose that a = b, then

ab		=	a²
∴	ab - b²	=	a² - b²
∴	b(a - b)	=	(a + b)(a - b)
∴ b		=	a + b
∴ b		=	2b
∴ 1		=	2

This process only proves in reality that 0 × 1 = 0 × 2, which is true.

No. Lvii.—Chess Cameo Double First Prize By A. Cyril Pearson

BLACK

WHITE

White to play, and mate in four moves.

Solution

Quick Calculation

Few people know a very singular but simple method of calculating rapidly how much any given number of pence a day amounts to in a year. The rule is this:—Set down the given number of pence as pounds; under this place its half, and under that the result of the number of original pence multiplied always by five. Take, for example, 7d a day:—

£7	0	0
3	10	0
	2	11
£10	12	11

The reason for this is evident as soon as we remember that the 365 days of a year may be split up into 240, 120, and 5, and that 240 happens to be the number of pence in a pound.

Science at Play

No. Lviii.—The Geared Wheels

A small wheel with ten teeth is geared into a large fixed wheel which has forty teeth. This small wheel, with an arrow mark on its highest cog, is revolved completely round the large wheel. How often during its course is the arrow pointing directly upwards? Here is a diagram of the starting position.

Solution

No. Lix.—Advancing Backwards

Here is a most curious and interesting question:—When an engine is drawing a train at full speed from York to London, what part of the train at any given moment is moving towards York?

At any time, when the engine is drawing a train at full speed from York to London, that part of the flange of each wheel which is for the moment at its lowest is actually moving backwards towards York.

For any point, such as A, on the circumference of the tyre, describes in running along a series of curves, as shown by full lines in the diagram; and any point, B, on the outer edge of the flange, follows a path shown by the dotted curves.

If these lines are followed round with a pencil in the direction of the arrows, it will be found that the point on the flange actually moves backwards as it passes below the track, while the point A, as it completes each curve, is at rest for the instant on the track, just before it starts afresh. The speed of the train does not affect these very curious facts.

No. Lx.—The Fifteen Bridges

In the subjoined diagram A and B represent two islands, round which a river runs as is indicated, with fifteen connecting bridges, that lead from the islands to the river’s banks.

Can you contrive to pass in turn over all these bridges without ever passing over the same one twice?

Solution

Arranging the Digits

In a school where two boys were taught to think out the bearings of their work, a sharp pupil remarked that 100 is represented on paper by the smallest digit and two cyphers, which are in themselves symbols of nothing. The master, quick to catch any signs of mental activity, took the opportunity to propound to his class the following ingenious puzzle:—How can the sum of 100 be represented exactly in figures and signs by making use of all the nine digits in their reverse order? This is how it is done:—

9 × 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 100.

Another ingenious method of using the nine digits, so that by simple addition they sum up to exactly 100, and each is used once only, is this:—

15 + 36 + 47 = 98 + 2 = 100

Here is another arrangement by which the nine digits written in their inverse order can be made to represent exactly 100:—

98 - 76 + 54 + 3 + 21 = 100.

Here is yet another way of arriving at 100 by using each of the digits, this time with an 0:—

40¹⁄₂
59³⁸⁄₇₆
100

No. Lxi.—Looping the Loop

Here is quite a pretty scientific experiment, which any one of a handy turn can construct and arrange:—

The spiral track is formed of two wires bent, and connected by curved cross-pieces. The upper twist is turned so that the ball starts on a horizontal course.

During the accelerated descent the ball acquires momentum enough to keep it on the vertical track, held outwardly against the wires by centrifugal force.

Convenient proportions are: height of spiral two feet, diameter six inches, and wire rails three-quarters of an inch apart.

No. Lxii.—A Mechanical Bird

A close approach to an ideal flying machine can be made with a little ingenuity. Two Y-shaped standards, secured to the backbone rod, support two wires which carry wings of thin silk, provided with light stays, and connected at their inner corners with the backbone by threads.

Rubber bands are attached to a loop on the inner end of the crank shaft, and secured to a post at the rear. These are twisted by turning the shaft with the cross wire, and when the tension is released the wings beat the air and carry the bird forward. It is known as Penaud’s mechanical bird, and has been sold as an attractive toy.

No. Lxiii.—Line of Swiftest Descent

A simple apparatus constructed on the lines of this illustration will give an interesting proof of the laws which govern falling bodies on an inclined plane or on a curved path.

In the case of the inclined plane the ball is governed by the usual law which controls falling bodies. In that of the concave circular curve, as it is accelerated rapidly at the start, it makes its longer journey in quicker time. In the case of the cycloidal curve it acquires a high velocity. This curve has therefore been called “the curve of swiftest descent,” as a falling body passes over it in less time than upon any path except the vertical.

No. Lxiv.—A Centrifugal Railway

Here is another very simple and pretty illustration of the natural forces which come into play in “looping the loop.”

This scientific toy on a small scale may be easily made, if care is taken that the height of the higher end of the rails is to the height of the circular part in a greater ratio than 5 to 4.

A ball started at the higher end follows the track throughout, and at one point is held by centrifugal force against the under side of the rails, against the force of gravity.

No. Lxv.—A Question of Gravity

If a ball is fired point blank from a perfectly horizontal gun, and travels half a mile over a level plain before it touches ground, and another similar ball is at the same moment dropped from the same height by some mechanical means, the two balls will touch ground simultaneously. The flight, however long, of one through the air has no influence upon the force of gravity, which draws it earthward at the same resistless rate as it draws the other that is merely dropped.

A Short Cut

A quick method of multiplying any number of figures by 5 is to divide them by 2, annexing a cypher to the result when there is no remainder, and if there is any remainder annexing a 5. Thus:—

464 × 5 = 2320; 464 ÷ 2 = 232, annex 0, = 2320.
753 × 5 = 3765; 753 ÷ 2 = 376, annex 5, = 3765.

No. Lxvi.—A Duck Hunt

A duck begins to swim round the edge of a circular pond, and at the same moment a water spaniel starts from the middle of the pond in pursuit of it.

If both swim at the same pace, how must the dog steer his course so that he is sure in any case to overtake the duck speedily?

Solution

The Magic of Dates

Louis Napoleon, Emperor 1852

	1852		1852		1852
date	1	date	1	date	1
of	8	of	8	of	8
his	0	Empress’s	2	their	5
birth	8	birth	6	marriage	3
	1869		1869		1869

Thus, by a most remarkable series of coincidences, the principal dates of the Emperor and Empress of the French added, as is shown above, to the year of the Emperor’s accession, express in each instance the year before his fall.

No. Lxvii.—Geometry with Dominoes

In this domino diagram we have a pretty and practical proof that the squares of the sides containing the right angle in any right-angled triangle are together equal to the square of the side opposite to the right angle.

Each stone forms two squares, and it is easily seen that the number of squares which make up the whole square on the line opposite the right angle are equal to the number of those which make up the two whole squares on the lines which contain that angle.

A second point to be noticed is that the number of pips on the large square are equal to the number on the other two squares combined, an arrangement of the stones which forms quite a game of patience to reproduce, if this pattern is not at hand.

No. Lxviii.—To Colour Maps

Four colours at most are needed to distinguish the surfaces of separate districts on any plane map, so that no two with a common boundary are tinted alike.

On this diagram A, B, and C, are adjoining districts, on a plane surface, and X borders, in one way or another, upon each.

It is clearly impossible to introduce a fifth area which shall so adjoin these four districts as to need another tint.

A Freak of Figures

Here is another freak of figures:—

9	×	1	-	1	=	8
9	×	21	-	1	=	188
9	×	321	-	1	=	2888
9	×	4321	-	1	=	38888
9	×	54321	-	1	=	488888
9	×	654321	-	1	=	5888888
9	×	7654321	-	1	=	68888888
9	×	87654321	-	1	=	788888888
9	×	987654321	-	1	=	8888888888

No. Lxix.—The Tethered Bird

A bird made fast to a pole six inches in diameter by a cord fifty feet long, in its flight first uncoils the cord, keeping it always taut, and then recoils it in the reverse direction, rewinding the coils close together. If it starts with the cord fully coiled, and continues its flight until it brings up against the pole, how far does it fly in its double course?

Solution

Strike It Out

Ask a person to write down in a line any number of figures, then to add them all together as units, and to subtract the result from the sum set down. Let him then strike out any one figure, and add the others together as units, telling you the result.

If this has been correctly done, the figure struck out can always be determined by deducting the final total from the multiple of 9 next above it. If the total happens to be a multiple of 9, then a 9 was struck out.

No. Lxx.—The Moving Disc and the Fly

A fly, starting from the point A, just outside a revolving disc, and always making straight for its mate at the point B, crosses the disc in four minutes, while the disc is revolving twice. What effect has the revolution of the disc on the path of the fly?

Solution

A Magic Square

This Magic Square is so arranged that the product of the continued multiplication of the numbers in each row, column, or diagonal is 4096, which is the cube of the central 16.

8	256	2
4	16	64
128	1	32

No. Lxxi.—A Shunting Puzzle

The railway, D E F, has two sidings, D B A and F C A, connected at A. The rails at A, common to both, are long enough to hold a single wagon such as P or Q, but too short to admit the whole of the engine R, which, if it runs up either siding, must return the same way.

How can the engine R be used to interchange the wagons P and Q without allowing any flying shunts?—From Ball’s Mathematical Recreations.

Solution

No. Lxxii.—A Curious Fact

It is a little known and very interesting fact that an equilateral triangle can easily be drawn by rule of thumb in the following way:—Take a triangle of any shape or size, and on each of its sides erect an equilateral triangle. Find and join the centres of these, and a fourth equilateral triangle is always thus formed, as shown by the dotted lines.

These centres are centres of gravity, and they are symmetrically distributed around the centre of gravity of the original triangle.

The figure formed by joining them must therefore be symmetrical, and, as in this case, it is a triangle, it must be always equilateral.

No. Lxxiii.—Try This Experiment

There can be no better instance of how the eye may be deceived than is so strikingly afforded in these very curious diagrams:—

The square which obviously contains sixty-four small squares, is to be cut into four parts, as is shown by the thicker lines. When these four pieces are quite simply put together, as shown in the second figure, there seem to be sixty-five squares instead of sixty-four.

This phenomena is due to the fact that the edges of the four pieces, which lie along the diagonal A B, do not exactly coincide in direction. In reality they include a very narrow diamond, not easily detected, whose area is just equal to that of one of the sixty-four small squares.

Figures in Swarms

Very curious are the results when the nine digits in reverse order are multiplied by 9 and its multiples up to 81. Thus:—

9	8	7	6	5	4	3	2	1	×	9	=		8	8	8	8	8	8	8	8	8	9
									×	18	=	1	7	7	7	7	7	7	7	7	7	8
									×	27	=	2	6	6	6	6	6	6	6	9	9	7
									×	36	=	3	5	5	5	5	5	5	5	5	5	6
									×	45	=	4	4	4	4	4	4	4	4	4	4	5
									×	54	=	5	3	3	3	3	3	3	3	3	3	4
									×	63	=	6	2	2	2	2	2	2	2	2	2	3
									×	72	=	7	1	1	1	1	1	1	1	1	1	2
									×	81	=	8	0	0	0	0	0	0	0	0	0	1

It will be seen that the figures by which the reversed digits are multiplied reappear at the beginning and end of each result except the first, and that the figures repeated between them are to be found by dividing the divisors by 9 and subtracting the result from 9. Thus, 54 ÷ 9 = 6, and 9 - 6 = 3.

No. Lxxiv.—A Triangle of Triangles

In this nest of triangles there are no less than six hundred and fifty-three distinct triangles of various shapes and sizes.

No. Lxxv.—Pharaoh’s Seal

In a chamber of the Great Pyramid an ancient Egyptian jar was found, marked with the device now known as Pharaoh’s seal.

Can you count the number of triangles or pyramids, of many sizes, but all of similar shape that are expressed on it? Solvers should draw the figure on a larger scale.

Solution

Making Cubes

It is interesting to note that the repeated addition of odd numbers to one another can be so arranged as to produce cube numbers in due sequence. Thus:—

1	=	1 × 1 × 1
3 + 5	=	2 × 2 × 2
7 + 9 + 11	=	3 × 3 × 3
13 + 15 + 17 + 19	=	4 × 4 × 4
21 + 23 + 25 + 27 + 29	=	5 × 5 × 5

and so on, to any extent.

No. Lxxvi.—Round the Garden

In a large old-fashioned garden walks were arranged round a central fountain in the shape of a Maltese cross.

If four persons started at noon from the fountain, walking round the four paths at two, three, four and five miles an hour respectively, at what time would they meet for the third time at their starting-point, if the distance on each track was one-third of a mile?

Solution

A Nice Short Cut

When the tens of two numbers are the same, and their units added together make ten, multiply the units together, increase one of the tens by unity, and multiply it by the other ten. The result is the product of the two original numbers, if the first result follows the other. Thus:—

43 × 47 = 2021.

No. Lxxvii.—A Joiner’s Puzzle

Can you cut Fig. A into two parts, and so rearrange these that they form either Fig. B or Fig. C?

The two parts of A must not be turned round to form B or C, but must retain their original direction.

Solution

A Calculation

Coal may fail us, but we can never run short of material for “words that burn.” It has been calculated that if a man could read 100,000 words in an hour, and there were 4,650,000 men available, they could not pronounce the possible variations which could be formed from the alphabet in 70,000 years!

A Paradox

It is possible, in a sense, by the following neat method, to take 45 from 45, and find that 45 remains:—

987654321	=	45.
123456789	=	45.
864197532	=	45.

No. Lxxviii.—The Broken Octagon

Cut out in stiff cardboard four pieces shaped as Fig. 1, four as Fig. 2, and four as Fig. 3, taking care that they are all exactly true to pattern in shape and proportion to one another.

Now see whether you can put the twelve pieces together so as to form a perfect octagon.

Solution

Properties of Seven

Here is a proof that 7, if it cannot rival the mystic 9, has quaint properties of its own:—

	1	5	8	7	3	×	7	=	1	1	1	1	1	1
	3	1	7	4	6	×	7	=	2	2	2	2	2	2
	4	7	6	1	9	×	7	=	3	3	3	3	3	3
	6	3	4	9	2	×	7	=	4	4	4	4	4	4
	7	9	3	6	5	×	7	=	5	5	5	5	5	5
	9	5	2	3	8	×	7	=	6	6	6	6	6	6
1	1	1	1	1	1	×	7	=	7	7	7	7	7	7
1	2	6	9	8	4	×	7	=	8	8	8	8	8	8
1	4	2	8	5	7	×	7	=	9	9	9	9	9	9

A Swarm of Eights

Here is an arithmetical curiosity:—

9	×	9								+	7	=	8	8
9	×	9	8							+	6	=	8	8	8
9	×	9	8	7						+	5	=	8	8	8	8
9	×	9	8	7	6					+	4	=	8	8	8	8	8
9	×	9	8	7	6	5				+	3	=	8	8	8	8	8	8
9	×	9	8	7	6	5	4			+	2	=	8	8	8	8	8	8	8
9	×	9	8	7	6	5	4	3		+	1	=	8	8	8	8	8	8	8	8
9	×	9	8	7	6	5	4	3	2	+	0	=	8	8	8	8	8	8	8	8	8

No. Lxxix.—At a Duck Pond

A farmer’s wife kept a pure strain of Aylesbury ducks for market on a square pond, with a duck-house at each corner. As trade grew brisk she found that she must enlarge her pond. An ingenious neighbour undertook to arrange this without altering the shape of the pond, and without disturbing the duck-houses. What was his plan?

Solution

Strange Subtraction

It would seem impossible to subtract 69 from 55, but it can be arranged thus, with six as a remainder:—

SIX	IX	XL
IX	X	L
S	I	X

No. Lxxx.—All on the Square

Cut out in cardboard twenty triangular pieces exactly the size and shape of this one, and try to place them together so that they form a perfect square.

Solution

Another Mystic Number

The decimal equivalent of ¹⁄₁₃ is.076923. This (omitting the point), multiplied by 1, 3, 4, 9, 10, or 12, yields results in which the same figures appear in varied order, but similar sequence, and multiplied by 2, 5, 6, 7, 8, or 11, it yields a different series, with similar characteristics. Thus:—

76923	×	1	=	76923	76923	×	2	=	153846
	×	3	=	230769		×	5	=	384615
	×	4	=	307692		×	6	=	461538
	×	9	=	692307		×	7	=	538461
	×	10	=	769230		×	8	=	615384
	×	12	=	923076		×	11	=	846153

Don’t Buy It to Try It

A kaleidoscope cylinder contains twenty small pieces of coloured glass. As we turn it round, or shake it, so as to make ten changes of pattern every minute, it will take the inconceivable space of time of 462,880,899,576 years and 360 days to exhaust all the possible symmetrical variations. (The 360 days is good!)

No. Lxxxi.—Pins and Dots

Here is an amusing little exercise for the ingenuity of our solvers.

Take six sharp pins, and puzzle out how to stick them into six of the black dots, so that no two pins, are on the same line, in any direction, vertical, horizontal, or diagonal.

Solution

No. Lxxxii.—A Tricky Course

The middle of a large playground was paved with sixty-four square flagstones of equal size, which are numbered on this diagram from one to sixty-four.

1	9	17	25	33	41	49	57
2	10	18	26	34	42	50	58
3	11	19	27	35	43	51	59
4	12	20	28	36	44	52	60
5	13	21	29	37	45	53	61
6	14	22	30	38	46	54	62
7	15	23	31	39	47	55	63
8	16	24	32	40	48	56	64

One of the schoolmasters, who had a head for puzzles, took his stand upon the square here numbered 19, and offered a prize to any boy who, starting from the square numbered 46, could make his way to him, passing through every square once, and only once. It was after many vain attempts that the course was at last discovered. Can you work it out?

Solution

No. Lxxxiii.—For the Children

Place twelve draughtsmen, or buttons, in a square, so that you count four along each side of it, thus:—

Now take the same men or buttons, and arrange them so that they form another square, and you can count five along each side of it.

Solution

A Game of Nines

Here is a good specimen of the eccentricities and powers of numbers:—

153846	×	13	=	1999998
230769	×	13	=	2999997
307692	×	13	=	3999996
384615	×	13	=	4999995
461538	×	13	=	5999994
538461	×	13	=	6999993
615384	×	13	=	7999992
692307	×	13	=	8999991

No. Lxxxiv.—Tell-Tale Tables

She was quite an old maid, and her age was a most absolute secret. Determined to discover it, her scapegrace nephew, on Christmas Eve, produced these tables, and asked her with well simulated innocence on which of them she could see the number of her age.

1	23	45	2	23	46	16	27	54
3	25	47	3	26	47	17	28	55
5	27	49	6	27	50	18	29	56
7	29	51	7	30	51	19	30	57
9	31	53	10	31	54	20	31	58
11	33	55	11	34	55	21	48	59
13	35	57	14	35	58	22	49	60
15	37	59	15	38	59	23	50	61
17	39	61	18	39	62	24	51	62
19	41	A	19	42	B	25	52	C
21	43	A	22	43	B	26	53	C

8	27	46	4	23	46	32	43	54
9	28	47	5	28	47	33	44	55
10	29	56	6	29	52	34	45	56
11	30	57	7	30	53	35	46	57
12	31	58	12	31	54	36	47	58
13	40	59	13	36	55	37	48	59
14	41	60	14	37	60	38	49	60
15	42	61	15	38	61	39	50	61
24	43	62	20	39	62	40	51	62
25	44	D	21	44	E	41	52	F
26	45	D	22	45	E	42	53	F

From her answer he was able to calculate that the old lady was fifty-five.

The tell-tale tables disclosed her age thus:—As it appeared in tables A, B, C, E, and F, he added together the numbers at the top left-hand corners, and found the total to be fifty-five. This rule applies in all cases.

No. Lxxxv.—A Paper Puzzle

Of the many paper-cutting tricks which appeal to us none is more simple and attractive than this:—

Take a piece of paper, say 5 inches by 3 inches, but any oblong shape and size will do, and after folding it four times cut it lengthways up the centre. Unfold the pieces, and to your surprise you will find a perfect cross and other pieces in pairs of the shapes shown above. The puzzle is how to fold the paper.

The paper must be folded first so that B comes upon C, then so that A comes upon D, then from D to C, and lastly from E to C. If it is now cut lengthways exactly along the centre the figures shown on the original diagram will be formed, which resemble a cross and lighted candles on an altar.

No. Lxxxvi.—A Home-Made Puzzle

Take a thin board, about eight inches square, and mark it out into thirty-six equal parts; bore a hole in the centre of each part, and then fit in a small wooden peg, leaving about a quarter inch above the surface, as is shown in Fig. 1, the section below the diagram.

Prepare thirty-six pieces of white or coloured cardboard of the length A to B, and place them over the pegs in any direction in which they will fit so as to form some such symmetrical pattern as is given on the second diagram, putting two holes only on each peg. Chess-players will see that this is the regular knight’s move.

Quite a number of beautiful designs can be thus formed, and those who have not the means at hand for making a complete set can enjoy the puzzle by merely marking out thirty-six squares, and drawing lines from centre to centre of the exact length from A to B, with black or coloured pencils.

No. Lxxxvii.—Loyd’s Mitre Problem

Divide this figure into four similar and equal parts.

Solution

A Pretty Problem

The solution of the pretty little problem: place three twos in three different groups, so that twice the first group, or half the third group equals the second group, is this:—

22 + 2 = 12 2 - 22 = 1 2 + 22 = 2

No. Lxxxviii.—Continuous Lines

The following figure, which represents part of a brick wall, cannot be marked out along all the edges of the bricks in less than six continuous lines without going more than once over the same line:—

Here, in strong contrast to the simple figure given above, which could not be traced without lifting the pen six times from the paper, is an intricate design, the lines of which, on the upper or on the lower half, can be traced without any break at all.

The general rule that governs such cases is, that where an uneven number of lines meet a fresh start has to be made. In the diagram now given the only such points are at the extremities of the upper and lower halves of the figure at A and X. At all other points two, or four, or six lines converge, and there is no break of continuity in a tracing of the figure.

No. Lxxxix.—Cut Off the Corners

Can you suggest quite a simple and practical way to fix the points on the sides of a square which will be at the angles of an octagon formed by cutting off equal corners of the square, as shown below?

Solution

Mystic Figures

Very interesting and curious are the properties of the figures 142857, used in varied order but always in similar sequence, in connection with 7 and 9:—

1	4	2	8	5	7	×	7	=		9	9	9	9	9	9	÷	9	=	1	1	1	1	1	1
2	8	5	7	1	4	×	7	=	1	9	9	9	9	9	8	÷	9	=	2	2	2	2	2	2
4	2	8	5	7	1	×	7	=	2	9	9	9	9	9	7	÷	9	=	3	3	3	3	3	3
5	7	1	4	2	8	×	7	=	3	9	9	9	9	9	6	÷	9	=	4	4	4	4	4	4
7	1	4	2	8	5	×	7	=	4	9	9	9	9	9	5	÷	9	=	5	5	5	5	5	5
8	5	7	1	4	2	×	7	=	5	9	9	9	9	9	4	÷	9	=	6	6	6	6	6	6

No. Xc.—The Five Triangles

The subjoined diagram shows how a square with sides that measure each 12 yards can be divided into five triangles, no two of which are of equal area, and of which the sides and areas can be expressed in yards by whole numbers:—

The areas of these triangles are 6, 12, 24, 48, and 54 square yards respectively, and the sum of these, 144 square yards, is the area of the square.

Curious Coincidences

Our readers may remember the remarkable fact that the figures of the sum, £12, 12s. 8d., when written thus, 12,128, exactly represent the number of farthings it contains. Now this, so far as we know, is the only instance of the peculiarity, but there are at least five other cases which come curiously near to it. They are these:—

£	s.	d.
9	9	6		=	9096	farthings
6	6	4		=	6064	„
3	3	2		=	3032	„
10	10	6	¹⁄₂	=	10106	„
13	13	8	¹⁄₂	=	13138	„

No. Xci.—Placing a Ladder

If a ladder, with rungs 1 foot apart, rests against a wall, and its thirteenth rung is 12 feet above the ground, the foot of the ladder is 25 feet from the wall.

Proof.—Drop a perpendicular from A to B. Then, as A B C is a right angle, and the squares on A C, A B, are 169 feet and 144 feet, the square on C B must be 25 feet, and the length of C B is 5 feet. We thus move 5 feet towards the wall in going 13 feet up the ladder, and in mounting 65 feet (five times as far) we must cover 25 feet.

No. Xcii.—Graceful Curves

A prettily ingenious method of dividing the area of a circle into quarters, each of them a perfect curve, with perimeter (or enclosing line) equal to the circumference of the circle, and with which four circles can be formed, is clearly shown by the subjoined diagrams:—

Nights at a Round Table

The host of a large hotel at Cairo noticed that his Visitors’ Book contained the names of an Austrian, a Brazilian, a Chinaman, a Dane, an Englishman, a Frenchman, a German, and a Hungarian. Moved by this curious alphabetical list, he offered them all free quarters and the best of everything if they could arrange themselves at a round dining-table so that not one of them should have the same two neighbours on any two occasions for 21 successive days.

The following is one of many ways in which this arrangement can be made, and it seems to be the simplest of them all.

Number the persons 1 to 8; and for our first day set them down in numerical order except that the two centre ones (4 and 5) change places:

(1st day)— 1 2 3 5 4 6 7 8

Keep the 1 and the 7 unaltered but double each of the other numbers. When the product is greater than 8, divide by 7, and only set down the remainder. Thus we get:

(8th day)— 1 4 6 3 8 5 7 2

(Here the fourth figure 3 is 5 × 2 ÷ 7, giving remainder 3, and so on.)

Repeat this operation once more:

(15th day)— 1 8 5 6 2 3 7 4

To fill in the intermediate days we have only to keep 1 unchanged and let the remaining numbers run downwards in simple numerical order, following 8 with 2, 2 with 3, and so on. Thus:—

1st day—	1	2	3	5	4	6	7	8
2nd day—	1	3	4	6	5	7	8	2
3rd day—	1	4	5	7	6	8	2	3
4th day—	1	5	6	8	7	2	3	4
5th day—	1	6	7	2	8	3	4	5
6th day—	1	7	8	3	2	4	5	6
7th day—	1	8	2	4	3	5	6	7
8th day—	1	4	6	3	8	5	7	2
9th day—	1	5	7	4	2	6	8	3
10th day—	1	6	8	5	3	7	2	4
11th day—	1	7	2	6	4	8	3	5
12th day—	1	8	3	7	5	2	4	6
13th day—	1	2	4	8	6	3	5	7
14th day—	1	3	5	2	7	4	6	8
15th day—	1	8	5	6	2	3	7	4
16th day—	1	2	6	7	3	4	8	5
17th day—	1	3	7	8	4	5	2	6
18th day—	1	4	8	2	5	6	3	7
19th day—	1	5	2	3	6	7	4	8
20th day—	1	6	3	4	7	8	5	2
21st day—	1	7	4	5	8	2	6	3

This completes the schedule. It will be found on examination that every number is between every pair of the other numbers once, and once only.

In order to reduce our first-day ring to exact numerical order we have only to interchange the numbers 4 and 5 throughout. The first three lines for example would then become:

1	2	3	4	5	6	7	8
1	3	5	6	4	7	8	2
1	5	4	7	6	8	2	3,	etc.

or, by putting letters for figures,

A	B	C	D	E	F	G	H
A	C	E	F	D	G	H	B
A	E	D	G	F	H	B	C,	etc.

An arrangement of the guests is thus arrived at for twenty-one successive days, so that not one of them has the same two neighbours on any two occasions.

No. Xciii.—Making Many Squares

Can you apply the two oblongs drawn below to the two concentric squares, so as to produce thirty-one perfect squares?

Solution

No. Xciv.—Cut Across

Take a piece of cardboard in the form of a Greek cross with arms, as shown here, and divide it by two straight cuts, so that the pieces when reunited form a perfect square.

Solution

No. Xcv.—A Pretty Puzzle

The diagrams which we give below show how a hollow square can be formed of the pieces of three-quarters of another square from which a corner has been cut away:—

No. Xcvi.—A Five-Fold Square

The subjoined diagram shows how a square of paper or cardboard may be cut into nine pieces which, when suitably arranged, form five perfect squares.

The Sociable Schoolgirls

On how many days can fifteen schoolgirls go out for a walk so arranged in rows of three, that no two are together more than once?

Fifteen schoolgirls can go out for a walk on seven days so arranged in rows of three that no two are together more than once.

It is said, on high authority, that there are no less than 15,567,522,000 different solutions to this problem. Here is one of them, given in Ball’s Mathematical Recreations, in which k stands for one of the girls, and a, b, c, d, e, f, g, in their modifications, for her companions on the seven different days:—

Sunday	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
ka₁a₂	kb₁b₂	kc₁c₂	kd₁d₂	ke₁e₂	kf₁f₂	kg₁g₂
b₁d₁f₁	a₁d₂e₂	a₁d₁e₁	a₂b₂c₂	a₂b₁c₁	a₁b₂c₁	a₁b₁c₂
b₂e₁g₁	a₂f₂g₂	a₂f₁g₁	a₁f₂g₁	a₁f₁g₂	a₂d₂e₁	a₂d₁e₂
c₁d₂g₂	c₁d₁g₁	b₁d₂f₂	b₁e₁g₂	b₂d₁f₂	b₁e₂g₁	b₂d₂f₁
c₂e₂f₂	c₂e₁f₁	b₂e₂g₂	c₁e₂f₁	c₂d₂g₁	c₂d₁g₂	c₁e₁f₂

It is an excellent game of patience, for those who have time and inclination, to place the figures 1 to 15 inclusive in seven such columns, so as to fulfil the conditions.

No. Xcvii.—The Three Crosses

It is possible from a Greek cross to cut off four equal pieces which, when put together, will form another Greek cross exactly half the size of the original, and by this process to leave a third Greek cross complete.

This is how to do it:—

Bisect C D at N, F G at O, K L at P, and B I at Q.

Join N H, O M, P A, Q E, intersecting at R, S, T, U.

Bisect A R at V, E S at W, T H at X, and M U at Y.

Join V Q, N Y, W P, O X, N W, V O, Q X, Y P.

Carefully cut out from the original Greek cross the newly-formed Greek cross, and the odd pieces from around it can be arranged to form another Greek cross.

No. Xcviii.—The Hammock

The greatest number of plane figures that can be formed by the union of ten straight lines is thirty-six.

The two equal lines at right angles are first drawn, and each is divided into eight equal parts. The other eight straight lines are then drawn from a to a, from b to b, and so on, until the hammock-shaped network of thirty-six plane figures is produced.

No. Xcix.—Cutting a Crumpet

It will be seen on the diagram below that seven straight vertical cuts with a table-knife will divide a crumpet into twenty-eight parts.

Balance the Scales

The nine digits can be so adjusted as to form an equation, or, if taken as weight, to balance the scales. Thus:—

9, 6¹⁄₂ = 3, 5, 7⁴⁄₈

True Stretches of Imagination

How large is the sea? This is a bold big question, and any possible answer involves a considerable stretch of the imagination. Here is a startling illustration of its vast volume:—

If the water of the sea could be gathered into a round column, reaching the 93,000,000 miles which separate us from the sun, the diameter of this column would be nearly two miles and a half!

It is perhaps even more difficult to realise that this mighty mass of waters could be dissipated in a few moments, if the column we have imagined could become ice, and if the entire heat of the sun could be concentrated upon it. All would be melted in one second of time, and converted into steam in eight seconds!

No. C.—To Make an Envelope

Many of our readers may be glad to know an easy way to make an envelope of any shape or size.

This diagram speaks for itself. When the lines A B, A C, D B, D C have been drawn, the corners of the rectangle, H E G F, are folded over, as shown by the dotted lines, after the corners have been rounded, and the margins touched with gum.

No. Ci.—Squaring an Oblong

The diagrams below will show how a piece of paper, 15 inches long and 3 inches wide, can be cut into five parts, and rearranged to form a perfect square.

A Plague of Blow-Flies

The following astounding calculation is answer enough to a question put by one of the authors of “Rejected Addresses:”—“Who filled the butchers’ shops with big blue flies?”

A pair of blow-flies can produce ten thousand eggs, which mature in a fortnight. If every egg hatches out, and there are equal numbers of either sex, which forthwith increase and multiply at the same rapid rate, and if their descendants do the like, so that all survive at the end of six months, it has been calculated that, if thirty-two would fill a cubic inch of space, the whole innumerable swarm would cover the globe, land and sea, half a mile deep everywhere.

No. Cii.—By Rule of Thumb

Here is quite a neat way to make an equilateral triangle without using compasses:—

Take a piece of paper exactly square, which we will call A B C D, fold it across the middle, so as to form the crease E F; unfold it, and fold it again so that the corner D falls upon the crease E F at G, and the angle at G is exactly divided. Again unfold the square, and from G draw the straight lines G C and G D. Then G C D is the equilateral triangle required.

The Value of a Frenchwoman

How can we be sure that the value of a Frenchwoman is just 1 franc 8 centimes?

We can be sure that the exact value of a Frenchwoman is 1 franc 8 centimes, for

	Two Frenchwomen	=	Deux Françaises.
	Deux Françaises	=	deux francs seize. (2 francs 16).
Therefore,	One Frenchwoman	=	1 franc 8!

No. Ciii.—Clearing the Wall

If a 52-feet ladder is set up so as just to clear a garden wall 12 feet high and 15 feet from the building, it will touch the house 48 feet from the ground.

Our diagram shows this, and also, by a dotted line, the only other possible position in which it could fulfil the conditions, if it were then of any practical use.

A Burden of Pins

If one pin could be dropped into a vessel this week, two the next, four the next, and so on, doubling each time for a year, the accumulated quantity would be 4,503,599,627,370,495, and their weight, if we reckon 200 pins to the ounce, would amount to 628,292,358 tons, a full load for 27,924 ships as large as the Great Eastern, whose capacity was 22,500 tons.

No. Civ.—Memories of Euclid

When at the signpost which said “To A 4 miles, to B 9 miles” on one arm, and on the other “To C 3 miles, to D —— miles,” and the boy whom I met could only tell me that the farm he worked at was equidistant from A, B, C, and D, and nearer to them than to the signpost, and that all the roads ran straight, I found, thanks to memories of Euclid, that I was 12 miles from D.

Since B A and D C intersect outside the circle at the signpost E,

therefore	A E × E B = C E × E D.
but	A E × E B = 4 × 9 = 36,
therefore	C E × E D = 36,
and	C E = 3, therefore E D = 12.

Q.E.D.

No. Cv.—A Transformation

This seems to be quite a poor attempt at a Maltese cross, but there is method in the madness of its make.

It is possible by two straight cuts to divide this uneven cross into four pieces which can be arranged together again so that they form a perfect square. Where must the cuts be made, and how are the four pieces rearranged?

Solution

Brevity Is the Soul of Wit

We all remember that splendidly terse message of success sent home to the authorities by Napier when he had conquered the armies of Scinde—“Peccavi!” (I have sinned).

History had an excellent opportunity for repeating itself when Admiral Dewey defeated the Spaniel fleet, for he might have conveyed the news of his victory by the one burning word—“Cantharides”—“The Spanish fly!”

No. Cvi—Shifting the Cells

In the diagram below a square is subdivided into twenty-five cells.

Can you, keeping always on the straight lines, cut this into four pieces, and arrange these as two perfect squares, in which every semicircle still occupies the upper half of its cell?

Solution

A Home-Made Microscope

The following very simple recipe for a home-made microscope has been suggested by a Fellow of the Royal Microscopical Society:—

Take a piece of black card, make a small pinhole in it, put it close to the eye, and look at some small object closely, such as the type of a newspaper. A very decided magnifying power will be shown thus.

No. Cvii.—In a Tangle

Where on this diagram must we place twenty-one pins or dots so that they fall into symmetrical design and form thirty rows, with three in each row?

Solution

On All Fours

Those who are fond of figures will find it a most interesting exercise to see how far they are able to represent every number, from one up to a hundred, by the use of four fours. Any of the usual signs and symbols of arithmetic may be brought into use. Here are a few instances of what may thus be done:—

3 = 4 + 4 + 44; 9 = 4 + 4 + 44;

36 = 4(4 + 4) + 4; 45 = 44 + 44;

52 = 44 + 4 + 4; 60 = 4 × 4 × 4 - 4.

No. Cviii.—Still a Square

This figure, which now forms a square, and the quarter of that square, can be so divided by two straight lines that its parts, separated and then reunited, form a perfect square. How is this done?

Solution

A Magic Square

Here we have arranged five rows of five cards each, so that no two similar cards are in the same lines. Counting the ace as eleven, each row, column, and diagonal adds up to exactly twenty-six.

After you have looked at this Magic Square, and set it out on the table, shuffle the cards, and try to re-arrange them so as to give the same results.

No. Cix.—A Transformation

Cut a square of paper or cardboard into seven such pieces as are marked in this diagram.

Can you rearrange them so that they form the figure 8.

Solution

A Secret Revealed

There are several ways in which strange juggling with figures and numbers is to be done, but none is more curious than this:—

Ask someone, whose age you do not know, to write down secretly the date and month of his birth in figures, to multiply this by 2, to add 5, to multiply by 50, to add his age last birthday and 365. He then hands you this total only from this you subtract 615. This reveals to you at a glance his age and birthday.

Thus, if he was born April 7 and is 23, 74 (the day and the month) × 2 = 148; 148 + 5 = 153; 153 × 50 = 7650; 7650 + 23 (his age) = 7673; and 7673 + 365 = 8038. If from this you subtract 615, you have 7423, which represents to you the seventh day of the fourth month, 23 years age! This rule works out correctly in all cases.

No. Cx.—To Make an Oblong

Cut out in stiff paper or cardboard two pieces of the shape and size of the small triangle, and four pieces of the shapes and sizes of the other three patterns—fourteen pieces in all.

The puzzle is to fit these pieces together so that they form a perfect oblong.

Solution

Sleepers That Slip and Sleep

Here is a string of sentences, which may be used as stimulating mental gymnastics when we leave the “Land of Nod.”

A sleeper runs on sleepers, and in this sleeper on sleepers sleepers sleep. As this sleeper carries its sleepers over the sleepers that are under the sleeper, a slack sleeper slips. This jars the sleeper and its sleepers, so that they slip and no longer sleep.

Days That Are Barred

Clever calculation has established a fact which we shall not be able to verify by personal experience. Whatever else may happen, the first day of a century can never fall on Sunday, Wednesday, or Friday.

No. Cxi.—Squares on the Cross

On this cross there are seventeen distinct and perfect squares marked out at their corners by asterisks.

How few, and which, of these can you remove, so that not a single perfect square remains?

Solution

Standing Room for All

Solution

3.

A traveller said to the landlord of an inn, “Give me as much money as I have in my hand, and I will spend sixpence with you.” This was done, and the process was twice repeated, when the traveller had no money left. How much had he at first?

Solution

4.

How can we obtain eleven by adding one-third of twelve to four-fifths of seven?

Solution

Fill in the Gaps

5. Can you replace the missing figures in this mutilated long division sum?

2	1	5	)	*	7	*	9	*	(	1	*	*
				*	*	*
				*	5	*	9
				*	5	*	5
						*	4	*
						*	*	*

Solution

6.

I buy as many heads of asparagus in the market as can be contained by a string 1 foot long. Next day I take a string 2 feet long, and buy as many as it will gird, offering double the price that I have given before. Was this a reasonable offer?

Solution

Alike from Either End

7. As “one good turn deserves another,” first reverse me, then reverse my square, then my cube, then my fourth power. When all this is done no change has been made. What am I?

Solution

The Sealed Bags

8. How can a thousand pounds be so conveniently stored in ten sealed bags that any sum in pounds from £1 to £1000 can be paid without breaking any of the seals?

Solution

9.

This is at once a problem and a puzzle:—

Though you twist and turn me over,
Yet no change can you discover.
Take me thrice, and cut in twain,
You will find but one remain.

Solution

10.

Three gamblers, when they sit down to play, agree that the loser shall always double the sum of money that the other two have before them. After each of them has lost once, it is found that each has eight sovereigns on the table. How much money had each at starting?

Solution

Find the Multiplier

11. When Tom’s back was turned, the boy sitting next to him rubbed out almost all his sum. Tom could not remember the multiplier, and only this remained on his slate—

		3	4	5
			.	.
	.	.	.	.
.	.	.	.
.	.	7	6	.

Can you reconstruct the sum?

Solution

Juggling with the Digits

12. The sum of the nine digits is 45. Can you hit upon other arrangements of 1, 2, 3, 4, 5, 6, 7, 8, 9, writing each of them once only, which will produce the same total. Of course fractions may be used.

Solution

A Brain Twister

13. The combined ages of Mary and Ann are forty-four years. 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, then, is Mary?

Solution

14.

Mr Oldboy was playing backgammon with his wife on the eve of his golden wedding, and could not make up his mind whether he should leave a blot where it could be taken up by an ace, or one which a tré would hit.

and the square roots of these:—

2 - ⁵⁄₂ = 3 - ⁵⁄₂

therefore 2 = 3.

Solution

Parcel Post Limitations

20. Our Parcel Post regulations limit the length of a parcel to 3 feet 6 inches, and the length and girth combined to 6 feet. What is the largest parcel of any size that can be sent through the post under these conditions?

Solution

21.

How can we show, or seem to show, that either four, five, or six nines amount to one hundred?

Solution

Test Your Skill

22. Can you arrange nine numbers in the nine cells of a square, the largest number 100, and the least 1, so that the product by multiplication of each row, column and diagonal is 1000?

Solution

Catching Crabs

23. Seven London boys were at the seaside for a week’s holiday, and during the six week-days they caught four fine crabs in pools under the rocks, when the tide was out at Beachy Head.

Hearing of this, the twenty-one boys of a school in the neighbourhood determined to explore the pools; but with the same rate of success they only caught one large crab. For how long were they busy searching under the seaweed?

Solution

Setting a Watch

24. On how many nights could a watch be set of a different trio from a company of fifteen soldiers, and how often on these terms could one of them, John Pipeclay, be included?

Solution

“Imperial Cæsar”

25. If Augustus Cæsar was born on September 23rd, B.C. 63, on what day and in what year did he celebrate his sixty-third birthday, and by what five-letter symbol can we express the date?

Solution

29. Three men gathered mangoes, and agreed that next day they would give one to their monkey and divide the rest equally. The first who arrived gave one to the monkey, and then took his proper share; the second came later and did likewise, and the third later still, neither knowing that any one had preceded him. Finally they met, and, as there were still mangoes, gave one to the monkey, and shared the rest equally. How many mangoes at least must there have been if all the divisions were accurate?

Solution

30.

I look at my watch between four and five, and again between seven and eight. The hands have, I find, exactly changed places, so that the hour-hand is where the minute-hand was, and the minute-hand takes the place of the hour-hand. At what time did I first look at my watch?

Solution

Three women, with no money, went to market. The first had thirty-three apples, the second twenty-nine, the third twenty-seven. Each woman sold the same number of apples for a penny. They all sold out, and yet each received an equal amount of money. How was this?

Solution

Simple Subtraction

38.

44.

If cash is lent at five per cent.
To those who choose to borrow,
How soon shall I be worth a pound
If I lend a crown to-morrow?

Solution

On the Jump

45. If I jump off a table with a 20-lb. dumb-bell in my hand, what is the pressure upon me of its weight while I am in the air?

Solution

At a Bazaar

46. In charitable mood I went recently to a bazaar where there were four tents arranged to tempt a purchaser. At the door of each tent I paid a shilling, and in each tent I spent half of the money remaining in my purse, giving the door-keepers each another shilling as I came out.

It took my last shilling to pay the fourth door-keeper. How much money had I at first, and what did I spend in each tent?

Solution

56. A fanciful collector, who bought pictures with more regard to quantity than quality, gave instructions that the area of each frame should exactly equal that of the picture it contained, and that the frames should be of the same width all round.

At an auction he picked up a so-called “old master,” unframed, which measured 18 inches by 12 inches. What width of frame will fulfil his conditions?

Solution

A Family Register

57. Our family consists of my mother, a brother, a sister, and myself. Our average age is thirty-nine. My mother was twenty when I was born; my sister is two years my junior, and my brother is four years younger than she is. What are our respective ages?

Solution

58.

A spider in a dockyard unwittingly attached her web to a mechanical capstan 1 foot in diameter, at the moment when it began to revolve. To hold her ground she paid out 73 feet of thread, when the capstan stopped, and she found herself drained of silk.

One-eighth were lain
On beds of pain,
With hundreds six beside;
One-fifth were dead,
Captives, or fled,
Lost in grim warfare’s tide.

76.

A herring and a half costs a penny and a half; what is the price of a dozen?

Solution

77.

What sum of money is in any sense seen to be the double of itself?

Solution

Comic Arithmetic

78. At the close of his lecture upon unknown quantities, Dr Bulbous Roots, in playful mood, wrote this puzzle on his blackboard:—

Divide my fifth by my first and you have my fourth; subtract my first from my fifth and you have my second; multiply my first by my fourth followed by my second, and you have my third; place my second after my first and you have my third multiplied by my fourth. What am I?

Solution

Dropped Through the Globe

79. If we can imagine the earth at a standstill for the purpose of our experiment, and if a perfectly straight tunnel could be bored through its centre from side to side, what would be the course of a cannon ball dropped into it from one end, under the action of gravity?

Solution

Love Letters

80.

A lady to her lover cried,
“How many notes have you of mine?”
“Six more I’ve sent,” the youth replied,
“Than I have had of thine.”

“But if from one pound ten you take
The pennies we on stamps have spent,
One eighth their cost you thus will make.”
How many had they had and sent?

Solution

81.

A man, on the day of his marriage, made his will, leaving his money thus:—If a son should be born, two-thirds of the estate to that son and one-third to the widow. If a daughter should be born, two-thirds to the widow and one-third to that daughter. In the course of time twins were born, a boy and a girl. The man fell sick and died without making a fresh will. How ought his estate to be divided in justice to the widow, son, and daughter?

Solution

82.

My carpet is 22 feet across. My stride, either backwards or forwards, is always 2 feet, and I make a stride every second. If I take three strides forwards and two backwards continuously until I cross the carpet, how long does it take me to reach the end of it?

Solution

86. A man met a beggar and gave him half the money he had in his pocket, and a shilling besides. Meeting another he gave him half of what was left and two shillings, and to a third, he gave half of the remainder and three shillings. This left a shilling in his pocket. How much had he at first?

Solution

87.

A young clerk wishes to start work at an office in the City on January 1st. He has two promising offers, one from A of £100 a year, with a yearly rise of £20, the other from B of £100 a year with a half-yearly rise of £5. Which should he accept, and why?

Solution

How Can I Pay My Bill?

88. I have an abundance of florins and half-crowns, but no other coins. In how many different ways can I pay my tailor £11, 10s. without receiving change?

Solution

89.

A monkey climbing up a greased pole ascends 3 feet and slips down 2 feet in alternate seconds till he reaches the top. If the pole is 60 feet high, how long does it take him to arrive there?

Solution

“Safe Bind, Safe Find”

90. Old Adze, the village carpenter, who kept his tools in an open chest, found that his neighbours sometimes borrowed and forgot to return them.

To guard against this, he secured the lid of the chest with a letter lock, which carried six revolving rings, each engraved with twelve different letters. What are the chances against any one discovering the secret word formed by a letter on each ring, which will open the lock, and be the only key to the puzzle?

Solution

97. “Now, boys!” said Dr Tripos, “I think of a number, add 3, divide by 2, add 8, multiply by 2, subtract 2, and thus arrive at twice the number I thought of.” What was it?

Solution

98.

Two club friends, A and B, deposit similar stakes with C, and agree that whoever first wins three games at billiards shall take the whole of them. A wins two games and B wins one. Upon this they determine to divide the stakes in proper shares. How must this division be arranged?

Solution

103. I bought a parcel of nuts at forty-nine for twopence. I divided it into two equal parts, one of which I sold at twenty-four, the other at twenty-five for a penny. I spent and received an integral number of pence, but bought the least possible number of nuts. How many did I buy? What did they cost? What did I gain?

Solution

Money Matters

104. My purse contained sovereigns and shillings. After I had spent half of its contents there were as many pounds left as I had shillings at first. With what sum did I start?

Solution

A Delicate Question

105. A lady was asked her age in a letter, and she replied by postcard thus:—

If first my age is multiplied by three,
And then of that two-sevenths tripled be,
The square root of two-ninths of this is four,
Now tell my age, or never see me more!

What was her age?

Solution

106.

If cars run at uniform speed on the twopenny tube, from Shepherd’s Bush to the Bank at intervals of two minutes, how many shall I meet in half an hour if I am travelling from the Bank to Shepherd’s Bush?

Solution

Payment by Results

107. What would it cost me to keep my word if I were to offer my greengrocer a farthing for every different group of ten apples he could select from a basket of a hundred apples?

Solution

108.

If the minute-hand of a clock moves round in the opposite direction to the hour-hand, what will be the real time between three and four, when the hands are exactly together?

Solution

109.

Two monkeys have stolen some filberts and some walnuts. As they begin their feast they see the owner of the garden coming with a stick. It will take him two and a half minutes to reach them. There are twice as many filberts as walnuts, and one monkey finishes the walnuts at the rate of fifteen a minute in four-fifths of the time and bolts. The other manages to finish the filberts just in time.

If every try had been converted they would have made four-fifths of the maximum which a score of 34 points from tries and goals can yield.

Squares in Streets

150. The square of the number of my house is equal to the difference of the squares of the numbers of my next door neighbour’s houses on either side.

My brother, who lives in the next street, can say the same of the number of his house, though his number is not the same as mine. How are our houses numbered?

Solution

151.

Two men of unequal strength are set to move a block of marble which weighs 270 lbs., using for the purpose a light plank 6 feet long. The stronger man can carry 180 lbs. How must the block be placed so as to allow him just that share of the weight?

Solution

152.

A man has twenty coins, of which some are shillings and the rest half-crowns. If he were to change the half-crowns for sixpences and the shillings for pence he would have 156 coins. How many shillings has he?

Solution

Count the Coins

153. Some coins are placed at equal distances apart on a table, so that they form the sides of an equilateral triangle.

How old was Nellie, and how old was her dad?

Solution

Word and Letter Puzzles

A Magic Cocoon

This Magic Cocoon is so cleverly spun that the word can be traced and read in many ways.

					N
				N	O	N
			N	O	O	O	N
		N	O	O	C	O	O	N
	N	O	O	C	O	C	O	O	N
N	O	O	C	O	C	O	C	O	O	N
	N	O	O	C	O	C	O	O	N
		N	O	O	C	O	O	N
			N	O	O	O	N
				N	O	N
					N

How many readings can you discover starting from one or other of the Cs, and passing up and down or sideways, but not diagonally, and never over the same letter twice in a reading? There are 756!

The Chronogram

The Chronogram, severely classed by Addison as “a species of false wit” is a sentence in which the salient letters represent in Roman numerals some particular year. A good English specimen is this: “My Day Closed Is In Immortality.” The capital letters in these words give MDCIII., or 1603, the year in which Queen Elizabeth died.

A French Chronogram

The battle-cry at Montlhéry in 1465 was:—“à CheVaL, à CheVaL, gendarMes, à CheVaL!” Taking the letters printed in capitals—

M	=	1000
CCC	=	300
LLL	=	150
VVV	=	15
we have the date of the battle		1465

A Notable Chronogram

QVI CHRISTI IAVDES CANTANT
SANCTÆ PASSIONIS SVÆ VIRTVTE
IN IPSO ET PATRE VNVM SINT.

This curious inscription is placed over the organ at Ober Ammergau. Add together its Roman numerals and they give the date at which the organ was dedicated. The English of it is:—

May those who sing the praises of Christ be by virtue of His Sacred Passion one in The Father and in Himself.

A Perfect Chronogram

On a damaged inscription to Bishop Berkeley in Winchester Cathedral are the words—VIXI, LVXI—I have lived, I have shone. Added together in their values as Roman numerals the letters of these two words give his age exactly at his death—eighty-three.

A Prize Motto

On the return of the C.I.V. from the Boer War a prize was offered by Truth for the best motto appropriate to them. This was to consist of three words of which the first must begin with C the second with I and the third with V.

The prize was taken by the following Latin motto which is singularly happy both in construction and in meaning:—

CIVI	IVI	VICI
I roused	I went	I won.

The sequence of events is perfect; no letters but C.I.V. are used and the motto is a palindrome if read by syllables.

The Musical Scale

As in olden days some of the Psalms and other writings were constructed in acrostic form, so in the Middle Ages even serious writers would juggle with letters, as though they felt that such tricky methods were an aid to memory.

It was in this spirit that Guido Aretino, a Benedictine monk of Tuscany in 1204, gave names to the notes used in the musical scale from the first syllables of the lines of a Latin hymn. “Ut” is still used in France, though we and the Italians have substituted “do.”

UT queant laxis	REsonare fibris,
MIra gestorum	FAmuli tuorum,
SOLve polutis	LAbii reatis
O Pater alme!

All the Alphabet!

Many of us know that there is a long verse in the Book of Ezra in which all the letters of the alphabet are used, taking “j” as “i” (Ezra vii., v. 21).

This very curious coincidence also occurs in a comparatively short sentence in “The Beth Book,” by Sarah Grand:—“It was an exquisitely deep blue just then, with filmy white clouds drawn up over it like gauze;” and here “j” is itself in evidence.

Apt Alliteration

Schopenhauer, the famous German philosopher, who was a confirmed bachelor and misogynist, was compelled while living at Frankfort to support an old lady who had been crippled by his violence. When her death came as a welcome relief to him, he composed the following clever epitaph:—

Obit anus,
Abit onus.

which by the interchange of two letters pictured the position. It may be freely rendered:—

Old lady dies,
My burden flies.

A Double Sequence

The following clever composition, which appeared in the pages of Truth, contains a double sequence of words, which increase a letter at a time, the same letters appearing in varied order until at last “o” culminates in thornless, and “a” in restrainest. It is quite a remarkable tour-de-force.

O lack-a-day! at eve we sat,
One star had lit its lamps on high.
We did not note the circling bat,
Start from the stone when flitting nigh.
For the strait gate of honest doubt
Shut off the thrones of Love and Gain;
We dreamed not, as we mourned without,
That Time’s swift transit shortens pain.
O Thou, Who trainest souls to shine,
Though once we craved a thornless lot,
This gracious truth we now divine:
The bruised reed Thou strainest not;
But by restraints, that gently tame;
Restrainest Passion’s kindling flame.

The Reign of Terror

During the Reign of Terror, France and her people and position were thus alphabetically described:—

Le peuple Français	A B C.	(abaissé).
La gloire nationale	F A C.	(effacée).
Les places fortes	O Q P.	(occupées).
Quarante trois députés	C D.	(cédés).
L’armée	D P C.	(dépaysée).
Les ministres	A J.	(agés).
La liberté	O T.	(ôtée).
La charte	L U D.	(éludée).

See-Saw

This elaborate method of piling up no less than seven consecutive “thats,” so that they make tolerable sense, was told to his boys during school-time by Dr Moberly, then headmaster of Winchester, and afterwards Bishop of Salisbury, just fifty years ago:—

I saw that C saw.

C saw that that I saw.

I saw that that that C saw was so.

C saw that, that that that I saw was so.

I saw that, that that that that C saw was so.

C saw that that, that that that that I saw was so.

I saw that that, that that that that that C saw was so.

For a Roman Holiday

If the Roman ladies and children, at their equivalent for Christmas, amused themselves by acting verbal charades, an excellent word was at their disposal, “sustineamus”—“let us endure,” which can be broken up exactly into sus, tinea, mus—a sow, a moth, a mouse.

A Word Square

1. Can you complete this word square, so that its four words read alike from top to bottom and from left to right?

*	E	*	*
E	*	*	*
*	*	*	E
*	*	E	*

Solution

Another Word Square

2. Can you fill in this word square?

C	*	*	C	*	E
*	N	U	*	E	S
*	U	*	E	S	*
C	*	E	*	S	*
*	E	S	*	*	E
E	*	T	E	*	*

Solution

Duplicate Letters

3. In this sentence, when complete,

So * * * * AG * * * * LATI * * * * X * * * * ITH

each group of four missing letters contains two pairs of letters which are alike. Can you on these lines complete the sentence?

Here is a similar sentence by way of illustration:

T * * * * M * * * * TERTAIN * * * * MUND,

which becomes when filled in—

Two women entertained Edmund.

Solution

Another Word Square

4. Can you complete this word square by substituting letters for the dots?

W	*	*	*	E
*	*	T	*	*
*	T	O	N	*
*	*	N	*	*
E	*	*	*	T

Solution

Word Building

5. What word can be made with these?

L S D U D O D U D.

Solution

6.

A lovelorn youth consulted a married lady on his condition, and was asked by her on a slip of paper:—

“Loruve?”

When he had deciphered this, and had answered in the affirmative, she handed to him another slip, on which this advice was written:—

		L
“Prove	A		F	and ensure success.”
		D

What did it all mean?

Solution

A Double Acrostic

Saint of Spain, whose daily word
Twenty years hath London heard!
Sweet days, elastic metal, motion sharp.
My halls once echoed to an Irish harp.
The “son of sorrow,” honourable of yore.
Dread goddess, loosing the loud dogs of war.
Time’s atom—total of eternities.
This name an insect bears, a patriot bore.
“So do, yet hear me,” said Themistocles.

Solution

Another Word Square

8. Can you complete this word square?

*	M	*	N	*	S
M	*	N	*	O	*
*	N	*	B	*	E
N	*	B	*	L	*
*	O	*	L	*	R
S	*	E	*	R	*

Solution

A French Oracle

9. A spruce young Frenchman at a fête consulted a modern oracle as to how he could best please the ladies. This was the mystic response:—

MEC DO BIC.

Can you interpret it?

Solution

A Quaint Equation

10. In our young days we have often wrestled with vulgar fractions, but apart from Algebra we have had no serious concern with any in which letters take the place of figures. A specimen of this sort, not known to science, is the following curiosity:—

m ot y = mo.

Solution

11.

The puzzle in Truth was recently founded upon “ourang-outang,” which had been cleverly buried. We will give a few of the best results. This is one:—

Poor wretch! a moisture filled his eye,
“Do not rebuff a lonely boy,”
Said he, “If ere I sink and die
Your smile—O! pardon will be joy!”

Another is:—

Though I jump, and row, and run,
Cap or cup I never won.

What animals are buried in these lines?

Solution

Like a Peacock’s Tail

12.

Fourteen letters here we fix,
Vowels only two are spoken;
All together these we mix
Into what can not be broken.

Solution

A Weird Word

13. There is an English word of thirteen letters in which the same vowel occurs four times, the same consonant six times, another consonant twice, and another once. Can you hit upon it?

Solution

A Condensed Proverb

14. Though brevity is said to be the soul of wit, we are too often flooded nowadays with a superabundance of words.

Here is an attempt at modest condensation. A familiar English proverb is quite clearly expressed to the solver’s seeing eyes in this brief phrase:—

WE IS DO

What is the proverb?

Solution

Another Word Square

15. Can you complete this word square?

W	*	*	*	*	S
*	R	*	*	R	*
*	*	O	R	*	*
*	*	R	M	*	*
*	R	*	*	N	*
S	*	*	*	*	M

Solution

Can You Decipher It?

16. The following puzzle lines are attributed to Dr Whewell:—

			O	O	N	O	O.
U	O	A	O	O	I	O	U
O	N	O	O	O	O	M	E	T	O	O.
U	O	A	O	I	D	O	S	O
I	O	N	O	O	I	O	U	T	O	O!

Solution

A Broken Diamond

17. Can you fill in the vacancies in this diamond?

				P
			F	O	*
		C	*	R	*	*
	F	*	*	C	*	*	*
P	O	R	C	E	L	A	I	N
	R	*	*	L	*	*	*
		S	*	A	*	*
			S	I	*
				N

Its words must read alike from left to right and from top to bottom.

Solution

What Is This?

18.

Tan HE Edsa VEN in
It N Gja SmeTs AsgN
aD Az Rett De.

Solution

A Phonetic Journey

19. I can travel first-class on the Great Eastern Railway from 2 2 2 2 2 2 2 2 4 4 4 4 4 5 0 0. What is the cost of my journey, and its length in time?

Solution

A Curious Old Inscription

20.

Seogeh sreve ereh wcisume vahl
Lah sehs se otreh nos llebdnas
Regni freh nos gnires rohyer
Ganoed iryd ale nifae esots sorcy
Rub nabot es rohk co caed ir.

Can you decipher it?

Solution

Irish Stew at Simpson’s

21. I wrote the following note recently:—

Dear Jack,—Meet me at Simpson’s to-morrow at 1.30. We will sample their excellent Irish stew. Here are some catchwords that will remind you of the invitation:—

Join me at and
Join me at ai
Join me at as

Why should they remind him of it?

Solution

English as She Is Spelt!

22. This was the exact text of a letter sent to the master of an English village school by a labourer as an excuse for his boy’s absence:—

“Cepatomtogoatatrin”

Can you decipher it?

Solution

A Double Acrostic

23. This double Acrostic will afford an easy exercise in mental gymnastics for those to whom such pastime appeals:—

Now we are fain
To rack your brain.

1. More fit for babes and sucklings than for you.

2. Robbed of externals this is very true.

3. Diminutive in measure and in weight.

4. Pen-name of one a true pen potentate.

5. A palindrome quite plain is here in sight.

6. Sans head and tail it also yields this light.

7. Here is in short what anyone may write.

Solution

Find the Proverb

24.	c	e	f	h	i	m	n	o	r	s	t	v	y
	3	2	2	2	7	1	1	2	6	5	8	3	9
		4		4	1		2	1		6	1
		3			1		5	8		2	3
		5			7		9				6
		4

The letters with ones under them are the first letters of words, those with twos under them are second letters of words, and so on.

Solution

A Puzzle Will

25. Having occasion to make a few slight additions to my will, I called in my lawyer to arrange the matter. How far forward did the instructions contained in the following lines carry him in his work?

Set down a hundred in my will,
Add nothing to the text;
Five hundred now a space may fill,
And one be added next.

Another hundred write as well,
And yet another one;
Then fifty more, and try to tell
The deed that now is done.

Solution

Another Word Square

26. Can you complete this word square?

*	D	*	*	O	*
D	*	*	I	*	E
*	S	*	A	*	D
T	*	A	*	A	*
*	R	*	A	*	E
R	*	D	*	E	*

Solution

Musical Epitaphs

27. Over the grave of a French musician, who was choked by a fish bone, the following epitaph was inscribed in notes of music:—A. G. A. E. A.

Over the porch of the house of Gustave Doré these musical notes were placed on a tablet:—C. E. B. A. C. D.

What do these inscriptions signify?

Solution

Quite Too Too

28. “Where can we meet to-morrow?” said Jack Spooner to his best girl.

“We will go,” she replied, “at 222222222222 LEY STREET.”

When and where did they meet?

Solution

A Broken Word

29. What does this spell?

CT T T T T T T T T T

Solution

Contradictory Terms

30. What English word is it which may be so treated as to affirm or disallow the use of its own initial or final letter?

Solution

Printers’ Pie

31. Can you arrange these letters

E I O O O U
B C N N R R S S

so that they form the title of a book well-known to boys?

Solution

35. Can you, by filling in letters, complete this word square so that it shall read alike across and from top to bottom?

*	A	*	*
A	*	*	A
*	E	*	*
*	A	*	E

Solution

A Quaint Inscription

36. The following curious inscription may be seen on a card hanging up in the bar of an old riverside inn in Norfolk:—

THEM * ILL * ERSLEA * VET * HEMI
LLT * HEW * HER * RYMEN * LOW
ERTH * EIRS * AILTH * EMA
LTS * TER * SLE * AVET * HE * KI
LN * FORAD * ROPO * FTH
EWHI * TESW * AN * SALE.

Can you decipher it?

Solution

A Poet’s Pi

37.

TONDEBNIOTOCHUMFOARYHUR
OTDIRECTTHAWHOTERSOFKLSYA;
TIKATESTUBALIGHTSTILLETRUFLYR
OTBOWLALLNEFESLEAVARFWYAA.

In this printer’s pie the words are in their proper sequence, but the letters are tangled.

Solution

Buried Places

38. In the following short sentences five names of places are buried—that is to say, the letters which spell them in proper order form parts of more words than one. Thus, for example “Paris” might be buried in the words “go up a rise:”

“The men could ride all on donkeys, the skipper, though, came to a bad end.”

When you have discovered these places, try to find out what very unexpected word of more than four letters is buried in the sentence, “On Christmas Eve you rang out angel peals.”

Solution

Treason Condoned

39. According to an old poet, Sir John Harrington (1561-1612):—

“Treason doth never flourish; what’s the reason?
For if it prosper none dare call it treason!”

The classic lines may possibly have been the germ of the flippant modern riddle, “Why is it no offence to conspire in the evening?”

Solution

A Bit of Botany

40.

Inscribe an m above a line
And write an e below,
This woodland flower is hung so fine
It bends when zephyrs flow.

Solution

A Pied Proverb

41. The following letters, if they are properly rearranged, will fall into the words which form a popular proverb:—

A A E E G G H I L L M N N N O O O O R R S S S S T T

Can you place them in position?

Solution

A Drop Letter Puzzle

42. Can you fill in the gaps of this proverb?

E**t* *e*s**s *a*e *h* *o** **i*e.

Solution

Multum in Parvo

43. There is an English word of five syllables which has only eight letters, five of them vowels—an a, an e, twice i, and y. What are its consonants?

Solution

Doublets

44. Can you turn TORMENT to RAPTURE, using four links, changing only one letter each time, and varying the order of the letters?

Solution

A Pied Proverb

45. Can you arrange these letters so that they form a sentence of five words?

a a c e e e f f h h i i i i i m n n o o o p r r s s t t t t t.

The result is a well-known English proverb.

Solution

What Can It Be?

46. Add one letter, and make this into a sensible English sentence:—

G D L D P R T F R R T H D X X F R D D N S

Solution

Out of Proportion

47.

One vowel in an English word is found,
Which by eight consonants is hedged around.

Solution

48.

Can you form an English word with these letters?

A A A A A B B N N I I R S S T T.

Solution

49.

What is this? It is found in Shakespeare:—

K I N I.

Solution

53. The names of eight famous British poets are buried in these lines, that is to say, the letters that spell the names form in their proper order parts of different words:—

The sun is darting rays of gold
Upon the moor, enchanting spot,
Whose purpled heights, by Ronald loved,
Up open to his Shepherd cot.

And sundry denizens of air
Are flying, aye, each to his nest;
And eager make at such an hour
All haste to reach the mansions-blest.

Can you dig them up?

Solution

A Fateful Letter

54. When A. B. gave up the reins of government, and C. B. took office in his place, it was found that their political positions could be exactly described by two quite common English verbs, which differ only in this, that the one is longer by one letter than the other, while the rest of the letters are the same, and in the same order. What are these two verbs?

Solution

A Prize Rebus

55. The following is a prize Rebus:—

Chess Cameos

No. XXVI

Black has made the false move Kt from Q sq to Kt 3. When this is replaced, and the king is moved as the proper penalty, White mates at once with one or other of the Knights.

Back to Cameo

No. XXVII

Replace the White Kt at B 7, and a Black Pawn at K 4; then P takes P en pas. Mate.

Analysis and Proof

It can be proved that Black’s last move must have been P from K 2 to K 4, so that White may take the P en pas.

The Black King cannot have moved from any occupied square.

(The White Kt now occupies B 7.)

Nor from Kt 3 or 4, as both are now doubly guarded, so that he cannot have moved out of a check.

(The White Kt now helps to guard Kt 5.)

Nor can he have moved from K 2, as the White P on Q 6 cannot have moved to give check.

No other P can have moved.

The K P cannot have moved from K 3, became of the position of the White King.

Therefore Black’s last move must have been P from K 2 to K 4, which White can take en pas giving Mate.

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No. XXVIII

B—Q 2P—R 7 B—R 5P—R 7 becomes Q P—Kt 4any White hasno move.

Kt—R 7B moves Q—KB 8B returns Q—R 8 mate.

Any other move of the Kt would impede the movements of the Q.

Kt—Kt 4, dis. ch.K—R 8. Q—KR 2, ch.P × Q. Kt—B 2, mate.

There are other variations.

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No. XLIII

B—Kt sq.P × Kt, Q—QR7,K × Kt. Q mates.

If K × Kt, Q × P, and mates next move.

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No. XLIV

K—Q 7,K moves R—Q 5,K × R Q mates.

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No. XLV

R—Q 8K moves Q × P, ch.K × Q B mates.

B moves Q—K 7any Q mates.

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No. XLVI

B—B 6K × R Q—Q 7, ch.K × Q R mates.

B × R R—B 6, ch.K × P Q mates.

There are other variations.

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R—QR sq.P moves R—R 2P × R P mates.

Back to Cameo

No. LII

This order of the bridges can, of course, be reversed.

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No. Lxvi.—A Duck Hunt

In order that a spaniel starting from the middle of a circular pond, and going at the same pace as a duck that is swimming round its edge, shall be sure to catch it speedily, the dog must always keep in the straight line between the duck and the centre of the pond.

The duck can never gain an advantage by turning back, and if it swims on continuously in a circle it will be overtaken when it has passed through a quarter of the circumference, for the dog will in the same time have described a semi-circle whose diameter is the radius of the pond, ending at the point where the duck is caught.

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No. Lxix.—The Tethered Bird

When a bird tethered by a cord 50 feet long to a post 6 inches in diameter uncoils the full length of the cord, and recoils it in the opposite direction, keeping it always taut, it flies 10,157 feet, or very nearly 2 miles, in its double course.

To avoid possible misunderstanding, we point out that, in order to pass from the uncoiling to the recoiling position, the bird must fly through a semicircle at the end of the fully extended cord.

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No. Lxx.—The Moving Disc and the Fly

When a fly, starting from the point A, just outside the revolving disc, and always making straight for its mate at the point B, crosses the disc in four minutes, during which time the disc is turning twice, the revolution of the disc has a most curious and interesting effect on the path of the fly.

The fly is a quarter of a minute in passing from the outside circle to the next, during which the disc has made an eighth of a revolution, and the fly has reached the point marked 1. The succeeding points up to 16 show the position of the fly at each quarter of a minute, until, by a prettily repeated curve, B is reached.

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No. Lxxi.—A Shunting Puzzle

The following method enables the engine R to interchange the positions of the wagons, P and Q, for either of which there is room on the straight rails at A, while there is not room there for the engine, which, if it runs up either siding, must return the same way:—

1. R pushes P into A. 2. R returns, pushes Q up to P in A, couples Q to P, draws them both out to F, and then pushes them to E. 3. P is now uncoupled, R takes Q back to A, and leaves it there. 4. R returns to P, pulls P back to C, and leaves it there. 5. R, running successively through F, D, B, comes to A, draws Q out, and leaves it at B.

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No. Lxxv.—Pharaoh’s Seal

It is quite puzzling to decide how many similar triangles or pyramids are expressed on the seal of Pharaoh. There are in fact 96.

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No. Lxxvi.—Round the Garden

The four persons who started at noon from the central fountain, and walked round the four paths at the rates of two, three, four, and five miles an hour would meet for the third time at their starting point at one o’clock, if the distance on each track was one-third of a mile.

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No. Lxxvii.—A Joiner’s Puzzle

This diagram shows how to divide Fig. A into two parts, and so rearrange these that they form either Fig. B or Fig. C, without turning either of the pieces.

Cut the five steps, and shift the two pieces as is shown.

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No. Lxxviii.—The Broken Octagon

The Broken Octagon is repaired and made perfect if its pieces are put together thus:—

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No. Lxxix.—At a Duck Pond

The pond was doubled in size without disturbing the duck-houses, thus:—

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No. Lxxx.—All on the Square

This is a perfect arrangement:—

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No. Lxxxi.—Pins and Dots

The pins may be placed thus:—

The following diagram shows by its dark lines how the whole square can be cut into four pieces, and these arranged as two perfect squares in which every semicircle still occupies the upper half of its cell.

One piece forms a square of nine cells, and it is easy to arrange the other three pieces in a square of sixteen cells by lifting the three cells and dropping the two.

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No. Cvii.—In a Tangle

It will be seen, on the subjoined diagram, how twenty-one counters or coins can be placed on the figure so that they fall into symmetrical design, and form thirty rows, with three in each row.

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No. Cviii.—Still a Square

In order that a square and an additional quarter may be divided by two straight lines so that their parts, separated and then reunited, form a perfect square, lines must be drawn from the point A to the corners B and C. Draw the figure on paper, cut through these lines, and you will find that the pieces can be so reunited that they form a perfect square.

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The dotted lines on the triangular figure show how a piece of cardboard cut to the shape of Fig. 1 can be divided into three pieces, and rearranged so that these form a star shaped as in Fig. 2.

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No. Cxiii.—Fireside Fun

To solve this puzzle slip the first coin or counter from A to D, then the others in turn from F to A, from C to F, from H to C, from E to H, from B to E, from G to B, and place the last on G. It can only be done by a sequence of this sort, in which each starting point is the finish of the next move.

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Amusing Problems

1. the Carpenter’s Puzzle

The carpenter cleverly contrived to mend a hole 2 feet wide and 12 feet long, by cutting the board which was 3 feet wide and 8 feet long, as is shown in Fig. 1, and putting the two pieces together as is shown in Fig. 2.

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2. Golden Pippins

Here is another solution:—

1	3	5	7	9	11	13	15	17	19
39	37	35	33	31	29	27	25	23	21
2	4	6	8	10	12	14	16	18	20
40	38	36	34	32	30	28	26	24	22
82	82	82	82	82	82	82	82	82	82

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3. an Awkward Fix

I was able to find my way in a strange district, when the sign-post lay uprooted in the ditch, without any difficulty. I simply replaced the post in its hole, so that the proper arm, with its lettering, pointed the way that I had come, and then, of necessity, the directions of the other arms were correct.

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4. Linked Sweetness Long Drawn Out

The train was whistling for 5 minutes. Sound travels about a mile in 5 seconds, so the first I heard of it was 5 seconds after it began. Its last sound reached me 7¹⁄₂ seconds after it ceased, so I heard the whistle for 5 minutes, 2¹⁄₂ seconds.

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5.

These quarters were not so elastic as they are made to appear. In good truth, considering that the second man who was placed in A was afterwards removed to I, no real second man was provided for at all.

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6.

The first day of a new century can never be Sunday, Wednesday, or Friday. The cycle of the Gregorian calendar is completed in 400 years, after which all dates repeat themselves.

As in this cycle there are only four first days of a century, it is clear that three of the seven days of the week must be excluded. Any perpetual calendar shows that the four which do occur are Monday, Tuesday, Thursday, and Saturday, so that Sunday, Wednesday, and Friday are shut out.

A neat corollary to this proof is that Monday is the only day which may be the first, or which may be the last, day of a century.

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7.

A cricket bat with spliced handle has such good driving power, because the elasticity of the handle allows the ball to be in contact with the blade of the bat for a longer time than would otherwise be possible.

With similar effect the “follow through” of the club head at golf maintains contact with the ball, when it is already travelling fast.

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8.

When two volumes stand in proper order on my bookshelf, each 2 inches thick over all, with covers ¹⁄₈ of an inch in thickness, a bookworm would only have to bore ¹⁄₄ of an inch, to penetrate from the first page of Vol. I, to the last page of Vol. II, for these pages would be in actual contact if there was no binding. This very pretty and puzzling question combines in its solution all the best qualities of a clever catch with solid and simple facts.

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9.

A man would have to fall from a height of nearly 15 miles to reach earth before the sound of his cry as he started. The velocity of sound is constant, while that of a falling body is continually accelerated. At first the cry far outstrips the falling man, but he overtakes and passes through his own scream in about 14¹⁄₂ miles, for his body falls through the 15 miles in 70 seconds, and sound travels as far in 72 seconds. Air resistance, and the fact that sound cannot pass from a rare to a dense atmosphere, are disregarded in this curious calculation.

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10.

A man on a perfectly smooth table in a vacuum, and where there was no friction, though no contortions of his body would avail to get away from this position, could escape from the predicament by throwing from him something which he could detach from his person, such as his watch or coat. He would himself instantly slide off in the opposite direction!

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11.

The monkey clinging to one end of a rope that passes over a single fixed pulley, while an equal weight hangs on the other end, cannot climb up the rope, or rise any higher from the ground.

If he continues to try to climb up, he will gradually pull the balancing weight on the other end of the rope upwards, and the slack of the rope will drop below him, while he remains in the same place.

If, after some efforts, he rests, he will sink lower and lower, until the weight reaches the pulley, because of the extra weight of rope on his side, if friction is disregarded.

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12.

Though the tension on a pair of traces tends as much to pull the horse backward as it does to pull the carriage forward, it is the initial pull from slack to taut which sets the traces in motion; and this, once started, must continue indefinitely until checked by a counter pull.

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13.

Some say that a rubber tyre leaves a double rut in dust and a single one in mud, because the air, rushing from each side into the wake of the wheel, piles up the loose dust. Others hold that the central ridge is caused by the continuous contraction of the tyre as it passes its point of contact with the road.

The initial letters of Turkey, Holland, England, France, Italy, Norway, Austria, Lapland, and Spain spell, and in this sense are the same as, “the finals.”

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Curious Calculations

1.

The only sum of money which satisfies the condition that its pounds, shillings, and pence written down as a continuous number, exactly give the number of farthings which it represents, is £12, 12s., 8d., for this sum contains 12,128 farthings.

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2.

If, when a train, on a level track, and running all the time at 30 miles an hour, slips a carriage which is uniformly retarded by brakes, and this comes to rest in 200 yards, the train itself will then have travelled 400 yards.

The slip carriage, uniformly retarded from 30 miles an hour to no miles an hour, has an average speed of 15 miles an hour, while the train itself, running on at 30 miles an hour all the time, has just double that speed, and so covers just twice the distance.

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3.

The traveller had fivepence farthing when he said to the landlord, “Give me as much as I have in my hand, and I will spend sixpence with you.” After repeating this process twice he had no money left.

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4.

This is the way to obtain eleven by adding one-third of twelve to four-fifths of seven—

TW(EL)VE + S(EVEN) = ELEVEN

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5.

Here is the completed sum:—

2	1	5	)	*	7	*	9	*	(	1	*	*	2	1	5	)	3	7	1	9	5	(	1	7	3
				*	*	*											2	1	5
				*	5	*	9										1	5	6	9
				*	5	*	5										1	5	0	5
						*	4	*											6	4	5
						*	*	*											6	4	5

The clue is that no figure but 3, when multiplied into 215, produces 4 in the tens place.

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6.

If I attempt to buy as many heads of asparagus as can be encircled by a string 2 feet long for double the price paid for as many as half that length will encompass, I shall not succeed. A circle double of another in circumference is also double in diameter, and its area is four times that of the other.

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7.

If, when you reverse me, and my square, and my cube, and my fourth power, you find that no changes have been made, I am 11, my square is 121, my cube 1331, and my fourth power 14641.

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8.

A thousand pounds can be stored in ten sealed bags, so that any sum in pounds up to £1,000 can be paid without breaking any of the seals, by placing in the bags 1, 2, 4, 8, 16, 32, 64, 128, 256, and 489 sovereigns.

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9.

It is the fraction ⁶⁄₉ which is unchanged when turned over, and which, when taken thrice, and then divided by two becomes 1.

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10.

When the three gamblers agreed that the loser should always double the sum of money that the other two had before them, and they each lost once, and fulfilled the conditions, remaining each with eight sovereigns in hand, they had started with £13, £7, and £4 as the following table shows:—

	A	B	C
	£	£	£
At starts	13	7	4
When A loses	2	14	8
When B loses	4	4	16
When C loses	8	8	8

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11.

Tom’s sum, which his mischievous neighbour rubbed almost out, is reconstructed thus:—

		3	4	5			3	4	5
			*	*				3	7
	*	*	*	*		2	4	1	5
*	*	*	*		1	0	3	5
*	*	7	6	*	1	2	7	6	5

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12.

Here are two other arrangements of the nine digits which produce 45, their sum; each is used once only:—

5 × 8 × 9 × (7 + 2)1 × 3 × 4 × 6 = 45

7² - 5 × 8 × 93 × 4 × 6 + 1 = 45

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13.

If, when the combined ages of Mary and Ann are 44, 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, Mary is 27¹⁄₂ years old, and Ann is 16¹⁄₂.

For, tracing the question backwards, when Ann was 5¹⁄₂ Mary was 16¹⁄₂. When Ann is three times that age she will be 49¹⁄₂. The half of this is 24³⁄₄, and when Mary was at that age Ann was 13³⁄₄. Mary’s age, by the question, was twice this, or 27¹⁄₂.

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14.

It is safer at backgammon to leave a blot in the tables which can be taken by an ace than one which a three would hit. In either the case of an actual ace or a three the chance is one in eleven; but there are two chances of throwing deuce-ace, the equivalent of three.

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15.

If I start from a bay, where the needle points due north, 1200 miles from the North Pole, and the course is perfectly clear, I can never reach it if I steam continuously 20 miles an hour, steering always north by the compass needle. After about 200 miles I come upon the Magnetic Pole, which so affects the needle that it no longer leads me northward, and I may have to steer south by it to reach the geographical Pole.

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16.

The 21 casks, 7 full, 7 half full, and 7 empty, were shared equally by A, B, and C, as follows:—

	Full cask.	Half full.	Empty.
A	2	3	2
B	2	3	2
C	3	1	3
or—
A	3	1	3
B	3	1	3
C	1	5	1

Thus each had 7 casks, and the equivalent of 3¹⁄₂ caskfuls of wine.

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17.

The foraging mouse, able to carry home three ears at a time from a box full of ears of corn, could not add more than fourteen ears of corn to its store in fourteen journeys, for it had each time to carry along two ears of its own.

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18.

If, with equal quantities of butter and lard, a small piece of butter is taken and mixed into all the lard, and if then a piece of this blend of similar size is put back into the butter, there will be in the end exactly as much lard in the butter as there is butter in the lard.

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19.

The fallacy of the equation—

4 - 10 = 9 - 15
4 - 10 + ²⁵⁄₄ = 9 - 15 + ²⁵⁄₄

and the square roots of these—

2 - ⁵⁄₂ = 3 - ⁵⁄₂
therefore 2 = 3

is explained thus:—The fallacy lies in ignoring the fact that the square roots are plus or minus. In the working we have taken both roots as plus. If we take one root plus, and the other minus, and add ⁵⁄₂, we have either 2 = 2, or 3 = 3.

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20.

The largest possible parcel which can be sent through the post under the official limits of 3 feet 6 inches in length, and 6 feet in length and girth combined, is a cylinder 2 feet long and 4 feet in circumference, the cubic contents of which are 2⁶⁄₁₁ cubic feet.

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21.

We can show, or seem to show, that either four, five, or six nines amount to 100, thus:—

99⁹⁄₉ = 100 IX
IX
IX
IX
IX
100 9 × 9 + 9 + 9⁹⁄₉ = 100

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22.

This is the magic square arrangement, so contrived that the products of the rows, columns, and diagonals are all 1,000.

50	1	20
4	10	25
5	100	2

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23.

If seven boys caught four crabs in the rock-pools at Beachy Head in six days, the twenty-one boys who searched under the seaweed and only caught one crab with the same rate of success were only at work for half a day.

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24.

A watch could be set of a different trio from a company of fifteen soldiers for 455 nights, and one of them, John Pipeclay, could be included ninety-one times.

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25.

If Augustus Cæsar was born September 23, B.C. 63, he celebrated his sixty-third birthday on September 23, B.C. 0; or, writing it otherwise, September 23, A.D. 0; or again, if we wish to include both symbols, B.C. 0 A.D. It is clear that his sixty-second birthday fell on September 23, B.C. 1, and his sixty-fourth on September 23, A.D. 1, so that the intervening year may be written as above.

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26.

The difference of the ages of A and B who were born in 1847 and 1874, is 27, or 30 - 03. Hence, when A was 30 B was 03. And A was 30 in 1877. Eleven years later A was 41 and B 14, and eleven years after that A was 52 and B 25. Thus the same two digits served to express the ages of both in 1877, 1888, and 1899. This can only happen in the cases of those whose ages differ by some multiple of nine.

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27

A hundred and one by fifty divide,
To this let a cypher be duly applied;
And when the result you can rightly divine,
You find that its value is just one in nine—

35.

The two numbers which solve the problem—

Two numbers seek which make eleven,
Divide the larger by the less,
The quotient is exactly seven,
As all who find them will confess—

are 1³⁄₈ and 9⁵⁄₈, for 1³⁄₈ + 9⁵⁄₈ = 11, and ⁷⁷⁄₈ ÷ ¹¹⁄₈ = 7.

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36.

There must be nine things of each sort, in order that 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 different selections may be made from twenty sorts of things.

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37.

The women who had respectively 33, 29, and 27 apples, and sold the same number for a penny, receiving an equal amount of money, began by selling at the rate of three a penny. The first sold ten pennyworth, the second eight pennyworth, and the third seven pennyworth.

The first had then left three apples, the second five, and the third six. These they sold at one penny each, so that they received on the whole—

The first	10d.	+	3d.	=	13d.
The second	8d.	+	5d.	=	13d.
The third	7d.	+	6d.	=	13d.

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38.

The puzzle—

Take five from five, oh, that is mean!
Take five from seven, and this is seen—

48.

When one monkey descends from the top of a tree 100 cubits high, and makes its way to a well 200 yards distant, while another monkey, leaping upwards from the top, descends by the hypotenuse to the well, both passing over an equal space, the second monkey springs 50 cubits into the air.

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49.

The steamboat which springs a leak 105 miles east of Tynemouth Lighthouse, and, putting back, goes at the rate of 10 miles an hour the first hour, but loses ground to the extent in each succeeding hour of one-tenth of her speed in the previous hour, never reaches the lighthouse, but goes down 5 miles short of it.

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50.

Twenty-one hens will lay ninety-eight eggs in a week, if a hen and a-half lays an egg and a-half in a day and a-half. Evidently one egg is laid in a day by a hen and a-half, that is to say three hens lay two eggs in a day. Therefore, twenty-one hens lay fourteen eggs, in a day, or ninety-eight in a week.

The arrangement by which St Peter is said to have secured safety for the fifteen Christians, when half of the vessel’s passengers were thrown overboard in a storm, is as follows:—

XXXXIIIIIXXIXXXIXIIXXIIIXIIXXI

Each Christian is represented by an X, and if every ninth man is taken until fifteen have been selected, no X becomes a victim.

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63.

If Farmer Southdown’s cow had a fine calf every year, and each of these, and their calves in their turn, at two years old followed this example, the result would be no less than 2584 head in sixteen years.

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64.

The number of the flock was 301. This is found by first taking the least common multiple of 2, 3, 4, 5, 6, which is 60, and then finding the lowest multiple of this, which with 1 added is divisible by 7. This 301 is exactly divisible by 7, but by the smaller numbers there is 1 as remainder.

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65.

The rule for determining easily the number of round bullets in a flat pyramid, with a base line of any length, is this:—

Add a half to half the number on the base line, and multiply the result by the number on that line. Thus, if there are twelve bullets as a foundation—

12 + ¹⁄₂ = ¹³⁄₂; and ¹³⁄₂ × ¹²⁄₁ = 78.

The same result is reached by multiplying the number on the base line by a number larger by one, and then halving the result. Thus—

12 × 13 = 156, 156 ÷ 2 = 78.

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66.

We can gather from the lines—

Old General Host
A battle lost,
And reckoned on a hissing,
When he saw plain
What men were slain,
And prisoners, and missing.

To his dismay
He learned next day
What havoc war had wrought;
He had, at most,
But half his host
Plus ten times three, six, ought.

One-eighth were lain
On beds of pain,
With hundreds six beside;
One-fifth were dead,
Captives, or fled,
Lost in grim warfare’s tide.

Now, if you can,
Tell me, my man,
What troops the general numbered,
When on that night
Before the fight
The deadly cannon slumbered?

that old General Host had an army 24,000 strong.

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67.

When the farmer sent five pieces of chain of 3 links each, to be made into one continuous length, agreeing to pay a penny for each link cut, and a penny for each link joined, the blacksmith, if he worked in the best interest of the farmer, could only charge sixpence: for he could cut asunder one set of 3 links, and use these three single links between the other four sets.

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68.

If, in a parcel of old silver and copper coins, each silver piece is worth as many pence as there are copper coins, and each copper coin is worth as many pence as there are silver coins, there are eighteen silver and six copper coins, when the whole parcel is worth eighteen shillings.

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69.

These are five groups that can be arranged with the numbers 1 to 11 inclusive, so that they are all equal:—

He sent 163. She sent 157.

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81.

When twins were born the estate was properly divided thus:—

Taking the daughter’s share as	1
The widow’s share would be	2
And the son’s share	4
Total	7	shares.

So the son takes four-sevenths, the widow two-sevenths, and the daughter one-seventh of the estate.

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82.

If each of my strides forwards or backwards across a 22 feet carpet is 2 feet, and I make a stride every second; and if I take three strides forwards and two backwards until I cross the carpet, I reach the end of it in forty-three seconds. In three steps I advance 6 feet. Then in two steps I retrace 4 feet, thus gaining only 2 feet in five steps, i.e., in five seconds. I therefore advance 16 feet in forty seconds, and three more strides cover the remaining 6 feet.

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83.

If the captain of a vessel chartered to sail from Lisbon to New York, which appear on a map of the world to be on the same parallel of latitude, and which are, along the parallel, about 3600 miles apart, takes his ship along this parallel, he will not be doing his best for the impatient merchant who has had an urgent business call to New York.

The shortest course between the two points is traced by a segment of a “great circle,” having its centre at the centre of the earth, and touching the two points. This segment lies wholly north of the parallel, and is the shortest possible course.

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84.

When John and Harry, starting from the right angle of a triangular field, run along its sides, and meet first in the middle of the opposite side, and again 32 yards from their starting point, if John’s speed is to Harry’s as 13 to 11, the sides of the field measure 384 yards.

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85.

If two sorts of wine when mixed in a flagon in equal parts cost 15d., but when mixed so that there are two parts of A to three of B cost 14d., a flagon of A would cost 20d., and a flagon of B 10d.

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Eight and a quarter is the answer to the nonsense question—

If five times four are thirty-three,
What will the fourth of twenty be?

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96.

The similar fraction of a pound, a shilling, and a penny which make up exactly a pound are as follows:—

		s.	d.
²⁴⁰⁄₂₅₃ of £1 =		18	11	¹⁶⁹⁄₂₅₃
²⁴⁰⁄₂₅₃ of 1s. =			11	⁹⁷⁄₂₅₃
²⁴⁰⁄₂₅₃ of 1d. =				²⁴⁰⁄₂₅₃
	£1	0	0

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97.

When Dr Tripos thought of a number, added 3, divided by 2, added 8, multiplied by 2, subtracted 2, and thus arrived at double the number, he started with 17.

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98.

When A and B deposited equal stakes with C, and agreed that the one who should first win three games of billiards should take all, but consented to a division in proper shares when A had won two games and B one, it was evident that if A won the next game all would go to him, while if he lost he would be entitled to one half. One case was as probable as the other, therefore he was entitled to half of these sums taken together; that is, to three quarters of the stakes, and B to a quarter only.

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99.

The average speed of a motor which runs over any course at 10 miles an hour, and returns over the same course at 15 miles an hour, is 12 miles an hour, and not 12¹⁄₂, as might be imagined. Thus a run of 60 miles out takes, under the conditions, six hours, and the return takes four hours; so that the double journey of 120 miles is done in ten hours, at an average speed of 12 miles an hour.

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100.

Farmer Hodge, who proposed to divide his sheep into two unequal parts, so that the larger part added to the square of the smaller part should equal the smaller part added to the square of the larger part, had but one sheep.

Faithful to his word, he divided this sheep into two unequal parts, ²⁄₃ and ¹⁄₃, and was able to show that ²⁄₃ + ¹⁄₉ = ⁷⁄₉, and that ¹⁄₃ + ⁴⁄₉ = ⁷⁄₉. He was heard to declare further, and he was absolutely right, that no number larger than 1 can be so divided as to satisfy the conditions which he had laid down.

The fact that sheep is both singular and plural, adds much to the perplexing points of this attractive problem.

Here is a very simple proof that the number must be 1:—

Let	a + b	=	no. of sheep
then	a² + b	=	b² + a
	a² - b²	=	a - b
or	(a + b)(a - b)	=	a - b
therefore	a + b	=	1.

Very Personal

168. When Nellie’s father said:—

I was twice as old as you are
The day that you were born.
You will be just what I was then
When fourteen years are gone—

he was 42, and she was 14.

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Word and Letter Puzzles

1.

The word square is completed thus:—

M	E	A	D
E	D	G	E
A	G	U	E
D	E	E	D

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2.

The word square filled in is:—

C	I	R	C	L	E
I	N	U	R	E	S
R	U	D	E	S	T
C	R	E	A	S	E
L	E	S	S	E	E
E	S	T	E	E	M

Notice the curious diagonal of E’s.

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3.

In the incomplete sentence,

SO * * * * AG * * * * LATI * * * * X * * * * ITH

the duplicate letters are filled in thus:—

SOME MEAGRE RELATIVE VEXED EDITH

The two last letters of each word are repeated as the two first of the word that follows.

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4.

The word square is completed thus:—

W	A	S	T	E
A	C	T	O	R
S	T	O	N	E
T	O	N	I	C
E	R	E	C	T

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5.

By twice building up two Ds into a B we make BULBOUS.

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6.

The question put on paper to the love-lorn youth, “Loruve?” is, when interpreted, “Are you in love?” and the advice given to him on another slip, “Prove L A FD and ensure success,” reads into, “Prove a fond lover, and ensure success” (a f on d l over).

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7. Double Acrostic St. James’ Gazette

S	prin	G
T	ar	A
J	abe	Z
A	t	E
M	omen	T
E	mme	T
S	trik	E

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8.

The completed word square is—

A	M	E	N	D	S
M	I	N	I	O	N
E	N	A	B	L	E
N	I	B	B	L	E
D	O	L	L	A	R
S	N	E	E	R	S

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9.

The oracular response to a young Frenchman at a fête, who inquired how he could best please the ladies—

MEC DO BIC

conceals this sage advice—

Aimez, cédez, obéissez!
Love, yield, obey!

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10.

The solution to our Letter Fraction Problem is of a verbal character. The original statement

m ot y = mo

is dealt with thus:—

m on ot on y = mo not on y, and so the word monotony solves the equation.

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11.

The buried beasts are chamois, buffalo, heifer, and leopard; and when the Oxford athlete cries—

“Though I jump and row and run,
Cap or cup I never won”

he introduces us to a porcupine.

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12.

The lines—

Fourteen letters here we fix,
Vowels only two are spoken;
All together these we mix
Into what can not be broken—

is solved by indivisibility, which has many an i, like a peacock’s tail.

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13.

The English word of thirteen letters in which the same vowel occurs four times, the same consonant six times, another twice, and another once, is Senselessness.

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14.

The condensed proverb “WE IS DO” reads at its full length as “Well begun is half done.”

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15.

This is the completed word square:—

W	A	S	H	E	S
A	R	T	E	R	Y
S	T	O	R	M	S
H	E	R	M	I	T
E	R	M	I	N	E
S	Y	S	T	E	M

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16.

Dr Whewell’s puzzle lines—

			O	O	N	O	O.
U	O	A	O	O	I	O	U
O	N	O	O	O	O	M	E	T	O	O.
U	O	A	O	I	D	O	S	O
I	O	N	O	O	I	O	U	T	O	O!

read thus:—

OH SIGH FOR NO CIPHER

You sigh for a cipher, O, I sigh for you,
Sigh for no cipher, O sigh for me too.
You sigh for a cipher, I decipher so,
I sigh for no cipher, I sigh for you too!

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17.

This is the completed diamond:—

				P
			P	O	R
		C	O	R	E	S
	F	O	R	C	E	P	S
P	O	R	C	E	L	A	I	N
	R	E	F	L	E	C	T
		S	P	A	C	E
			S	I	T
				N

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18.

The medley—

Tan HE Edsa VEN in
It N Gja SmeTs AsgN
aD Az Rett De

is read by taking first the capitals in their order, and then the small type. It comes put out as “The Evening Standard and Saint James’s Gazette.”

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19.

The statement, I can travel first-class on the G.E.R. from 2222222244444500, reads into—from 22 to 2 to 22 to 4 for 44 4d; or, in plain terms, from 1.38 to 3.38 for 14s. 8d. This works out at about 3d. a mile, the usual allowance for first-class, for two hours, at about 29 miles an hour.

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20. a Curious Old Inscription

Read the inscription backwards, and it resolves itself into the lines familiar to us in our childhood:—

Ride a cock horse to Banbury Cross
To see a fine lady ride on a grey horse.
Rings on her fingers and bells on her toes,
She shall have music wherever she goes!

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21.

The reason why the invitation to Jack to sample the Irish stew at Simpson’s was to be kept in mind by the catch words—

Join me at and
Join me at ai
Join me at as

is because if you join me and at, and note that and is on i, which is on s, you arrive at the suggestive sentence—meat and onions.

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22.

The labourer’s quaint letter, which ran “Cepatomtogoatatrin,” was, in plainer English, “Kept at home to go a tatering.”

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23.

Our double acrostic comes out thus:—

Problem—Puzzles

P	a	P
(T) R		U (E)
O		Z
B	o	Z
L	eve	L
E	v	E
M		S

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24.

The Hidden Proverb is—

“Necessity is the mother of invention.”

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25.

The “deed done” in our Will puzzle is the making in Roman numerals of “Codicil.” The lawyer was to set down, a hundred, to add nothing, to set down five hundred, then one, then another hundred, and then one more, and, finally, fifty, and accordingly he wrote upon the parchment the one word CODICIL.

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26.

The word square is—

E	D	I	T	O	R
D	E	S	I	R	E
I	S	L	A	N	D
T	I	A	R	A	S
O	R	N	A	T	E
R	E	D	S	E	A

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27.

The notes of music A. G. A. E. A. over the grave of a French musician, who was choked by a fish bone, are in the French notation, “La sol la mi la,” which reads into “La sole l’a mis 1à.”

Similarly the inscription over the porch of Gustave Doré’s house C. E. B. A. C. D. is equivalent to “Do, mi, si, la, do, re,” which may be taken to represent “Domicile à Doré.”

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28.

When his best girl said to Jack Spooner, “We can go to-morrow at 222222222222 LEY STREET,” he understood her to mean, “We can go to-morrow, at two minutes to two, to two twenty-two, to 222 Tooley Street.”

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29

CT T T T T T T T T T spells contents (c on ten ts!).

The quotation from Shakespeare—

OXXU8 MAAULGIHCTE

NOR

is by interpretation:—“Nothing extenuate, nor set down aught in malice.”

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34.

The phonetic nightmare—

Ieukngheaurrhphthewempeighghteaps—

is merely the word unfortunates. It can be justified thus by English spelling of similar sound taken letter by letter:—

u—iew in view; n—kn in know; f—gh in tough; o—eau in beau; r—rrh in myrrh; t—phth in phthisis; u—ewe; n—mp in comptroller; a—eigh in neigh; t—ght in light; e—ea in tea; and s—ps in psalm.

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35.

The word square is completed thus:—

F	A	R	M
A	R	E	A
R	E	N	T
M	A	T	E

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36. a Quaint Inscription

The millers leave the mill,
The wherrymen lower their sail;
The maltsters leave the kiln,
For a drop of the White Swan’s ale.

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37. a Poet’s Pi

TONDEBNIOTOCHUMFOARYHUR
OTDIRECTTHAWHOTERSOFKLSYA;
TIKATESTUBALIGHTSTILLETRUFLYR
OTBOWLALLNFESLEAVARFWYAA

is disentangled thus:—

Don’t be in too much of a hurry
To credit what other folks say;
It takes but a slight little flurry
To blow fallen leaves far away.

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38. Buried Places

The five buried places are Deal, London, Esk, Perth, and Baden. The word is Ourangoutang.

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39.

It is no offence to conspire in the evening, because what is treasonable is reasonable after t!

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40.

The bit of botany—

Inscribe an m above a line,
And write an e below,
This woodland flower is hung so fine
It bends when zephyrs blow—

is solved by, me, an em on e, the delicately hung wind-flower.

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41. a Pied Proverb

The pied proverb, is “A rolling stone gathers no moss.”

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42.

The Drop Letter Proverb—

47

One vowel in an English word is found,
Which by eight consonants is hedged around—

is solved by Strengths.

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48.

The letters AAAAABBNNIIRSSTT form the word Antisabbatarians.

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49.

The quotation from Shakespeare,

KINI

First a c and a t, last a c and a t,
With a couple of letters between,
Form a sight that our eyes are delighted to see,
Unless in their sight it is seen—

is solved by Cataract. The first line reads, First a c and a t, last a c and a t, that is cat and act.

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57.

Cuba.

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58.

The Rebus T S is solved by the words tones and tans, t before one s, or t before an s.

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59.

The phonetic phrase—

INXINXIN—

is, Ink sinks in!

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A Good End

60. Finis (F IN IS).

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