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.
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?
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?
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?
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?
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.
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.
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?
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.
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.
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?
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.
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.
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?
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.
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 | 22 | 43 | 26 | 53 |
| 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 | 22 | 45 | 42 | 53 |
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.
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?
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?
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.
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?
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?
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?
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?
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.
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.
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?
STANDING ROOM FOR ALL
On a globe 2 feet in diameter the Dead Sea appears but as a small coloured dot. If it were frozen over there would be standing room on its surface for the whole human race, allowing 6 square feet for each person; and if they were all suddenly engulfed, it would merely raise the level of the lake about 4 inches.
No. CXII.—A CHINESE PUZZLE
Here is quite a good exercise for ingenious brains and fingers. Cut a piece of stiff paper or cardboard into such a right-angled triangle as is shown below.
Can you divide this into only three pieces, which, when rearranged, will form the design given as No. 2?
A DEAL IN ROSES
I sent an order for dwarf roses to a famous nursery-garden, asking for a parcel of less than 100 plants, and stipulating that if I planted them 3 in a row there should be 1 over; if 4 in a row 2 over; if 5 in a row 3 over, and if 6 in a row 4 over, as a condition for payment.
The nurseryman was equal to the occasion, charged me for 58 trees, and duly received his cheque.
No. CXIII.—FIRESIDE FUN
This puzzle is not so easy of solution as it may seem at first sight.
Take a counter, or a coin, and place it on one of the points; then push it across to the opposite point, and leave it there. Do this with a second counter or coin, starting on a vacant point, and continue this process until every point is covered, as we place the eighth counter or coin on the last point.
A NOVEL EXERCISE
“Write down,” said a schoolmaster, “the nine digits in such order that the first three shall be one third of the last three, and the central three the result of subtracting the first three from the last.”
The arrangement which satisfies these conditions is, 219, 438, 657.