THE TWENTIETH CENTURY
Standard Puzzle
Book

[I-1]

In Albert Dürer’s day, as in Milton’s, “melancholy” meant thoughtfulness, and on this ground we find on his woodcut, “Melancholia, or the Genius of the Industrial Science of Mechanics,” a very early instance of a Magic Square, showing that Puzzles had a recognised place in mental gymnastics four hundred years ago.

[I-2]

Much time was devoted in olden days to the construction and elaboration of Magic Squares. Before we go more deeply into this fascinating subject, let us study the following pretty and ingenious method of making a Magic Square of sixteen numbers, which is comparatively simple, and easily committed to memory:—

Start with the small square at the top left-hand corner, placing there the 1; then count continuously from left to right, square by square, but only insert those numbers which fall upon the diagonals—namely, 4, 6, 7, 10, 11, 13, and 16.

Then start afresh at the bottom right-hand corner, calling it 1, and fill up the remaining squares in order, from right to left, counting continuously, and so placing in their turn 2, 3, 5, 8, 9, 12, 14, and 15. Each row, column, diagonal, and almost every cluster of four has 34 as the sum of its numbers.

[I-3]

In this Magic Square the rows, columns, and diagonals add up to 65, and the sum of any two opposite and corresponding squares is 26.

Solution

[I-4]

The numbers in this Magic Square of 49 cells add up in all rows, columns, and diagonals to 175. The four corner cells of every square or rectangle that has cell 25 in its centre, and cells 1, 7, 49, 43, add up to 100.

Solution

[I-5]

An ideal Magic Square can be constructed thus:

Place 1, 2, 3, 4, 5 in any order in the five top cells, set an asterisk over the third column, as shown in the diagram; begin the next row with this figure, and let the rest follow in the original sequence; continue this method with the other three rows.

Make a similar square of 25 cells with 0, 5, 10, 15, 20, as is shown in No. 2, placing the asterisk in this case over the fourth column of cells, and proceeding as before, in an unchanging sequence. Using these two preparatory squares, try to form a Magic Square in which the same number can be counted up in forty-two different ways.

Solution

[I-6]

Here is one of many methods by which a Magic Square of the first twenty-five numbers can readily be made.

This is done by first placing the figures from 1 to 25 in diagonal rows, as is shown above, and then introducing the numbers that are outside the square into it, by moving each of them five places right, left, up, or down. A Magic Square is thus formed, the numbers of which add up to 65 in lines, columns and diagonals, and with the centre and any four corresponding numbers on the borders.

[I-7]

Here is what may indeed be called a Champion Magic Square:—

Its 484 cells form, as they are numbered, a Magic Square, in which all rows, columns, and diagonals add up to 5335, and it is no easy matter to determine in how many other symmetrical ways its key-number can be found.

When the cells outside each of the dark border lines are removed, three other perfect Magic Squares remain.

Collectors should take particular note of this masterpiece.

[I-8]

A Magic Square of nine cells can be built up by taking any number divisible by 3, and placing, as a start, its third in the central cell. Thus:—

Say that 81 is chosen for the key number. Place 27 in the centre; 28, 29, in cells 1, 2; 30 in cell 7; 31 in 6; and then fill up cells 3, 4, 8, and 9 with the numbers necessary to make up 81 in each row, column, and diagonal.

Any number above 14 that is divisible by 3 can be dealt with in this way.

Solution

[I-9]

Among the infinite number of Magic Squares which can be constructed, it would be difficult to find a more remarkable setting of the numbers 1 to 32 inclusive than this, in which two squares, each of 16 cells, are perfect twins in characteristics and curious combinations.

There are at least forty-eight different ways in which 66 is the sum of four of these numbers. Besides the usual rows, columns, and diagonals, any square group of four, both corner sets, all opposite pairs on the outer cells, and each set of corresponding cells next to the corners, add up exactly to 66.

Solution

[I-10]

Here is a notable specimen of a Magic Square:—

The rows, columns, and diagonals all add up to exactly 175 in the full square. Strip off the outside cells all around, and a second Magic Square remains, which adds up in all such ways to 125.

Strip off another border, as is again indicated by the darker lines, and a third Magic Square is left, which adds up to 75.

Solution

[I-11]

Here is another example of what is called a “bordered” Magic Square:—

These 81 cells form a complete magic square, in which rows, columns, and diagonals add up to 369. As each border is removed fresh Magic Squares are formed, of which the distinctive numbers are 287, 205, and 123. The central 41 is in every case the greatest common divisor.

[I-12]

Can you complete this Magic Square, so that the rows, columns, and diagonals add up in every case to 505?

We have given you a substantial start, and, as a further hint, as all the numbers in the first and last columns end in 0 or 1, so in the two next columns all end in 2 or 9, in the two next in 3 or 8, in the two next in 4 or 7, and in the two central columns in 5 or 6.

Solution

Solution

[I-13]

In this Magic Square, not only do the rows, columns, and diagonals add up to 260, but this same number is produced in three other and quite unusual ways:—

(1) Each group of 8 numbers, ranged in a circle round the centre; there are six of these, of which the smallest is 22, 28, 38, 44, 19, 29, 35, 45, and the largest is 8, 10, 56, 58, 1, 15, 49, 63. (2) The sum of the 4 central numbers and 4 corners. (3) The diagonal cross of 4 numbers in the middle of the board.

[I-14]

On another page we give an interesting Magic Square of 121 cells based upon the figures of the year 1892. Here, in much more condensed form, is one more up to date.

The rows, columns, and diagonals of these nine cells add up in all cases to the figures of the year 1902.

The central 634 is found by dividing 1902 by its lowest factor greater than 2, and this is taken as the middle term of nine numbers, which are thus arranged to form a Magic Square.

Where does the treason come in?

Solution

[I-15]

The following square of numbers is interesting in connection with the year 1906.

and the sum, in every case, is 1906. [I-16]

Here is quite a curious nest of clustered Magic Squares, which is worth preserving:—

Every square of every possible combination of 25 of these numbers in their cells, such as the two with darker borders, is a perfect Magic Square, with rows, columns, and diagonals that add up in all cases to 65.

Solution

[I-17]

A very interesting method of constructing a Magic Square is shown in these three diagrams:—

It will be noticed that each row after the first, in the two upper auxiliary squares, begins with a number from the same column in the row above it, and maintains the same sequence of numbers. When the corresponding cells of these two squares are added together, and placed in the third square, a Magic Square is formed, in which 671 is the sum of all rows, columns, and diagonals.

[I-18]

Here is a curious form of Magic Square. The year 1892 is taken as its basis.

Within this square 1892 can be counted up in all the usual ways, and altogether in 44 variations. Thus any two rows that run parallel to a diagonal, and have between them eleven cells, add up to this number, if they are on opposite sides of the diagonal.

Solution

[I-19]

This Magic Square of 49 cells is constructed with a diagonal arrangement of the numbers from 1 to 49 in their proper order. Those that fall outside the central square are written into it in the seventh cell inwards from where they stand. It is interesting to find out the many combinations in which the number 175 is made up.

Solution

[I-20]

These are two interesting Magic Squares found on an antique gong, at Caius College, Cambridge:—

In the one nine numbers are so arranged that they count up to 27 in every direction; and in the other the outer rows total 30, while the central rows and diagonals make 40.

Solution

[I-21]

Among Moorish Mussulmans 78 is a mystic number.

Here is a cleverly-constructed Magic Square, to which this number is the key.

The number 78 can be arrived at in twenty-three different combinations—namely, ten rows, columns, or diagonals; four corner squares of four cells; one central square of four cells; the four corner cells; two sets of corresponding diagonal cells next to the corners; and two sets of central cells on the top and bottom rows, and on the outside columns.

[I-22]

Here is a Magic Square of singular charm:—

The 81 cells of this remarkable square are divided by parallel lines into 9 equal parts, each made up of 9 consecutive numbers, and each a Magic Square in itself within the parent square. Readers can work out for themselves the combinations in the larger square and in the little ones.

Solution

[I-23]

Fill each square by repeating two of its figures in the vacant cells. Then rearrange them all, so that the sums of the corresponding rows in each square are equal, and the sums of the squares of the corresponding cells of these rows are also equal; and so that the sums of the four diagonals are equal, and the sum of the squares of the cells in corresponding diagonals are equal.

Solution

There is an old-world charm about this Enigma:—

Solution

[I-24]

Here is an arrangement of fractions which form a perfect Magic Square:—

If these fractions are added together in any one of the eight directions, the result in every case is unity. Thus ³⁄₈ + ¹⁄₃ + ⁷⁄₂₄ = 1, ¹⁄₆ + ¹⁄₃ + ¹⁄₂ = 1, and so on throughout the rows, columns, and diagonals.

Solution

[I-25]

We are indebted to a friend for the following elaborate Magic Square of fractions, on the lines of that on the preceding page.

The composer claims that there are at least 160 combinations of 5 cells in which these fractions add up to unity, including, of course, the usual rows, columns, and diagonals.

Solution

[I-26]

On similar lines to Magic Squares, but as a distinct variety, we give below a specimen of a Magic Oblong.

The four rows of this Oblong add up in each case to 132, and its eight columns to 66. Two of its diagonals, from 10 to 5 and from 28 to 23, also total 66, as do the four squares at the right-hand ends of the top and bottom double rows.

Solution

Solution

[I-27]

Much more complicated than the Magic Square is the Magic Cube.

Lowest Layer.

Those who enjoy such feats with figures will find it interesting to work out the many ways in which, when the layers are placed one upon another, and form a cube, the number 315 is obtained by adding together the cell-numbers that lie in lines in the length, breadth, and thickness of the cube.

Solution

[I-28]

The Magic Circle below has this particular property:—

The 14 numbers ranged in smaller circles within its circumference are such that the sum of the squares of any adjacent two of them is equal to the sum of the squares of the pair diametrically opposite.

Solution

[I-29]

We have had some good specimens of Magic Squares. Here is a very curious and most interesting Magic Circle, in which particular numbers, from 12 to 75 inclusive, are arranged in 8 concentric circular spaces and in 8 radiating lines, with the central 12 common to them all.

The sum of all the numbers in any of the concentric circular spaces, with the 12, is 360, which is the number of degrees in a circle.

The sum of the numbers in each radiating line with the 12, is also 360.

The sum of the numbers in the upper or lower half of any of the circular spaces, with half of 12, is 180, the degrees of a semi-circle.

The sum of any outer or inner four of the numbers on the radiating lines, with the half of 12, is also 180.

[I-30]

In the following triangle, if two couples of the figures on opposite sides are transposed, the sums of the sides become equal, and also the sums of the squares of the numbers that lie along the sides. Which are the figures that must be transposed?

Solution

Solution

Solution

[I-31]

Here is a nest of concentric triangles. Can you arrange the first 18 numbers at their angles, and at the centres of their sides, so that they count 19, 38, or 57 in many ways, down, across, or along some angles?

This curiosity is found in an old document of the Mathematical Society of Spitalfields, dated 1717.

Solution

Solution

[I-32]

The numbers outside these twin triangles give the sum of the squares of the four figures of the adjacent sides:—

The twins are also closely allied on these points:—

18 is the common difference of 99, 117, 135, and of 119, 137, 155.

19 is the sum of each side of the upper triangle.

20 is the common difference of any two sums of squares symmetrically placed, both being on a line through the central spot.

21 is the sum of each side of the lower triangle.

10 is the sum of any two figures in the two triangles that correspond.

254 is the sum of 135, 119, of 117, 137, and of 90, 155.

By transposing in each triangle the figures joined by dotted lines, the nine digits run in natural sequence.

[I-33]

We have dealt with Magic Squares, Circles, and Triangles. Here is a Magic Hexagon, or a nest of Hexagons, in which the numbers from 1 to 73 are arranged about the common centre 37.

Each of these Hexagons always gives the same sum, when counted along the six sides, or along the six diameters which join its corners, or along the six which are at right angles to its sides. These sums are 259, 185, and 111.

Solution

[I-34]

Inscribe six equilateral triangles in a circle, as shown in this diagram, so as to form a regular hexagon.

Now place the nine digits round the sides of each of the triangles, so that their sum on each side may be 20, and so that, while there are no two triangles exactly alike in arrangement, the squares of the sums on the other sides may be alternately equal.

Solution

Solution

[I-35]

There are 33 different combinations of four of the numbers in the cells of this magic cross which total up in each case to 26.

Those who care to work them out on separate crosses will find that there is a very regular correspondence in the positions which the numbers occupy.

Solution

Solution

[I-36]

Here is quite a charming little puzzle, which is by no means easy of accomplishment:—

Start from one of these nine dots, and without taking the pen from the paper draw four straight lines which pass through them all. Each line, after the first, must start where the preceding one ends.

Solution

Solution

[I-37]

Place on a chess or draught-board three white men on the squares marked a, and three black men on the squares marked b.

The pieces marked a can only move one square at a time, from left to right, and those marked b one square at a time, from right to left, on to unoccupied squares; and any piece can leap over one of the other colour, on to an unoccupied square. What is the least number of moves in which the positions of the white and the black men can be reversed, so that each square now occupied by a white is occupied by a black, and each now occupied by a black holds a white piece?

Solution

Solution

Solution

[I-38]

In the upper row of this diagram four white and four black counters are placed alternately.

It is possible, by moving these counters two at a time, to arrange them in four moves as they stand on the lower row. Can you do this? Draughtsmen are handy for solving this puzzle, on a paper ruled as above.

Solution

Solution

Solution

[I-39]

Take five wooden matches, and bend each of them into a V. Place them together, as is shown in the diagram, so that they take the form of an asterisk, or a ten-pointed star.

Lay them on some smooth surface, and without touching them transform them into a star with five points.

Solution

Solution

Solution

[I-40]

It is possible, with plenty of patience, to build up a whole set of dominoes, so that they are safely supported on only two stones set up on end.

This, which might well seem impossible, is done by placing, as a foundation, dominoes in the positions indicated by dotted lines. The arch is then carefully constructed, as shown in the diagram, and for the finish the four stones between the two foundation arches are drawn out, and placed in pairs on end above, and finally, with the utmost care, the other four are drawn away, and built in on the top. Thus the stones indicated by the dotted lines at the base take their place within the dotted lines above.

[I-41]

This diagram represents a shallow box, on the bottom of which twelve counters or draughtsmen are lying loose.

How can they be readjusted so that they will wedge themselves together, and against the side of the box, and it can be turned upside down without displacing them?

Solution

Solution

[I-42]

The diagram below is an exact reproduction of an old-fashioned maze, cut in the ground near Nottingham. It is eighteen yards square, and the black line represents the pathway, which is 535 feet in length.

The point of this convoluted path is not so much to puzzle people, as to show how much ground may be covered without diverging far from a centre, or going over the same ground twice. As we advance along the line there are no obstructions, and we find ourselves, after passing over the whole of it, on the spot whence we set out.

Solution

[I-43]

Start at A, and trace these figures with one continuous line, finishing at B.

You must not take your pencil from the paper, or go over any line twice.

Solution

Solution

Solution

Solution

[I-44]

Those who have not seen it will find some real fun in the following little experiment. Fix three matches as shown in the diagram, light the cross match in the middle, and watch to see which of the ends will first catch fire, or what will happen.

Solution

Solution

[I-45]

Equal volumes of alcohol and water, when mixed, occupy less space than when separate, to the extent indicated in this picture:

If the sum of the volume of the two separate liquids is 100, the volume of the mixture will be only 94. It is thought that the molecules of the two liquids accommodate themselves to each other, so as to reduce the pores and diminish the volume of the mixture.

Solution

Solution

[I-46]

The simplest and cheapest of all microscopes can easily be made at home. The only materials needed are a thin slip of glass, on to which one or two short paper tubes, coated with black sealing wax, are cemented with the wax, a small stick, and a tumbler half full of water.

Water is dropped gradually by aid of the stick into the cells, until lenses are formed of the desired convexity, and objects held below the glass will be more or less magnified.

Solution

[I-47]

For this curious experiment a glass bottle or decanter about half full of water and a sound stalk of straw are needed.

Bend the straw without breaking it, and put it, as is shown, into the bottle, which can then be lifted steadily and safely by the straw, if it is a sound one.

A Russian nobleman had three sons. Rab, the eldest, became a lawyer, his brother Mary was a soldier, and the youngest was sent to sea. What was his name?

Solution

[I-48]

The button in a clear glass bottle, as is shown below, hangs attached by a thread to the cork, which is securely sealed at the top.

How can you sever the thread so that the button falls to the bottom without uncorking or breaking the bottle?

Solution

Solution

[I-49]

Here is a pretty trick which requires an empty bottle, a lucifer match, and a small coin.

Break the wooden match almost in half, and place it and the coin in the position shown above. Now consider how you can cause the coin to drop into the bottle, if no one touches it, or the match, or the bottle.

Solution

Solution

Solution

[I-50]

Our belief that two portions of matter cannot occupy the same space at the same time is almost shaken by the following experiment:

If we introduce slowly some fine powdered sugar into a tumblerful of warm water a considerable quantity may be dissolved in the water without increasing its bulk.

It is thought that the atoms of the water are so disposed as to receive the sugar between them, as a scuttle filled with coal might accommodate a quantity of sand.

Solution

[I-51]

Set a stool, as is shown in the diagram below, about nine or ten inches from the wall.

Clasp it firmly by its two side edges, plant your feet well away from it, and rest your head against the wall. Now lift the stool, and then try, without moving your feet, to recover an upright position.

It will be as impossible as it is to stand on one leg while the foot of that leg rests sideways against a wall or door.

[I-52]

Place a set of billiard balls as is shown in the diagram, the spot ball overhanging a corner pocket, and the red and the plain white in a straight line with it, leaving an eighth of an inch between the balls.

How can you pot the spot white with the plain white, using a cue, and without touching, or in any way disturbing, the red ball? There is not room to pass on either side between the red ball and the cushion.

Solution

[I-53]

Prepare a circular band of stiff paper, as is shown in the diagram, and balance it, with a coin on the top, on the lip of a bottle.

How can you most effectively transfer the coin into the bottle?

Solution

Solution

[I-54]

Place a strip of thin board, or a long wide flat ruler, on the edge of a table, so that it just balances itself, and spread over it an ordinary newspaper, as is shown in the illustration.

You may now hit it quite hard with your doubled fist, or with a stick, and the newspaper will hold it down, and remain as firmly in its place as if it were glued to the table over it. You are more likely to break the stick with which you strike than to displace the strip of wood or the paper. Try the experiment.

Solution

[I-55]

Here is an after-dinner balancing trick, which it is well to practise with something less brittle than the best glass:—

It will be seen that the blades of the knives are so cunningly interlaced as to form quite a firm support.

Solution

[I-56]

Here is a pretty little experiment, which shows the effect of liquids of different densities.

Drop an egg into a glass vessel half full of water, it sinks to the bottom. Drop it into strong brine, it floats. Introduce the brine through a long funnel at the bottom of the pure water, and the water and the egg will be lifted, so that the egg floats between the water and the brine in equilibrium. The egg is denser than the water, and the brine is denser than the egg.

Solution

[I-57]

Here is a very simple and effective little trick. Offer to balance an egg on its end on the lip of a glass bottle.

The picture shows how it is done, with the aid of a cork and a couple of silver forks.

(From “La Science Amusante”).

Solution

[I-58]

The boy in this picture is blowing hard against the bottle, which is between his mouth and the candle flame.

What will happen?

Solution

Solution

[I-59]

Here is a simple way to make a needle float on water:—

Fill a wineglass or tumbler with water, and on this lay quite flat a cigarette paper; place a needle gently on this, and presently the paper will sink, and the needle will float on the water.

Solution

[I-60]

Here is a pile of ten draughtsmen—one black among nine white.

If I take another draughtsman, and with a strong pull of my finger send it spinning against the column, what will happen?

Solution

Solution

Solution

[I-61]

How long would it take to divide completely a 2 ft. block of ice by means of a piece of wire on which a weight of 5 lb. hangs?

Solution

Solution

Solution

[I-62]

When weak cords of equal strength are attached to opposite parts of a wooden or metal ball which is suspended by one of them, a sharp, sudden pull will snap the lower cord before the movement has time to affect the ball; but a gentle, steady pull will cause the upper cord to snap, as it supports the weight below it.

Solution

[I-63]

Hold a pair of dice, and a cup for casting them, in one hand as is shown in the diagram.

Now, holding the cup fast, throw up one of the dice and catch it in the cup. How can you best be sure of catching the other also in the cup?

Solution

Here is a metrical Enigma, which appeals with particular force to all married folk, and to our cousins in America:

Solution

[I-64]

Build up seven dominoes into a double arch, as is shown in the diagram below, and place a single domino in the position indicated.

Now put the fore-finger carefully through the lower archway, and give this domino quite a smart tip up by pressing on its corner. What will happen if this is done cleverly? Try it.

Solution

Solution

[I-65]

Place three pennies in contact in a line as is shown below, so that a “head” is between two “tails.”

Can you introduce the coin with a shaded surface between the other two in a straight line, without touching one of these two, and without moving the other?

Solution

Solution

Solution

[I-66]

Place ten coins in a circle, as is shown in this diagram, so that on all of them the king’s head is uppermost.

Now start from any coin you choose, calling it 1, the next 2, and so on, and turn the fourth, so that the tail is uppermost. Start again on any king’s head, and again turn the fourth, and continue to do this until all but one are turned.

Coins already turned are reckoned in the counting, but the count of “four” must fall on an unturned coin.

Can you find a plan for turning all the coins but one in this way without ever failing to count four upon a fresh spot, and to start on an unturned coin?

Solution

[I-67]

The diagram below shows how, using one hand only, and grasping a bottle of wine by its body, the contents can be poured out without cutting or boring the cork, or altogether removing it from the bottle.

Solution

Solution

[I-68]

A sharply-pointed knife with a heavy handle is stuck very lightly into the lintel of a door, and the nut that is to be cracked is placed under it, so that when the knife is released by a touch the nut is cracked.

What simple and certain plan can you suggest for making sure that the knife shall hit the nut exactly in the middle without fail?

Solution

Solution

[I-69]

If we throw an ordinary wine cork into a tub of water it will naturally float on its side. It is, however, possible to arrange a group of seven such corks, without fastening them in any way, so that they will float in upright positions.

Place them together, as is shown in the illustration, and, holding them firmly, dip them under the water till they are well wetted. Then, keeping them exactly upright, leave go quietly, and they will float in a compact bunch if they are brought slowly to the surface.

Solution

[I-70]

As an exercise of patience and dexterity, try to balance a set of dominoes upon one that stands upon its narrow end:—

This is no easy matter, but a little patience will enable us to arrange the stones in layers, which can with care be lifted into place and balanced there.

Solution

[I-71]

We expect to puzzle our readers completely by this diagram:—

It is simply the enlargement by photography of part of a familiar picture.

Solution

Solution

Solution

[I-72]

Place a strip of smooth paper on a table so that it overhangs the side, as is shown in the diagram. Stand a new penny steadily on edge upon the paper.

Take hold of the paper firmly, and give it a smart, steady pull. If this is properly done it will leave the penny standing unmoved in its place.

Solution

Solution

[I-73]

Two men, standing on the bank of a broad stream, across which they could not cast their fishing lines, could not agree as to its width. A bet on the point was offered and accepted, and the question was presently decided for them by an ingenious friend who came along, without any particular appliances for measurement.

He stood on the edge of the bank, steadied his chin with one hand, and with the other tilted his cap till its peak just cut the top of the opposite bank.

Then, turning round, he stood exactly where the peak cut the level ground behind him, and, by stepping to that spot, was able to measure a distance equal to the width of the stream.

Solution

[I-74]

Paste or pin together the ends of a long strip of stiff paper so as to form a hoop, and place on the table a board resting at one end upon a book. Challenge those in your company to make the hoop run up the board without any impulse.

They must of course fail, but you can succeed by secretly fastening with beeswax a small stone or piece of metal inside the hoop, as is indicated in the diagram.

Solution

[I-75]

This double crescent, called the Seal of Mahomet, from a legend that the prophet was wont to describe it on the ground with one stroke of his scimitar, is to be made by one continuous stroke of pen or pencil, without going twice over any part of it.

Solution

Solution

Solution

[I-76]

Arrange 15 matches thus—

Remove 6 and what number will be left?

Solution

Solution

Solution

Can you discover ten fruits in these lines?

Solution

[I-77]

Solution

Solution

[I-78]

By the following ingenious arrangement of the stones a set of dominoes appears to be unduly rich in doublets:—

It will be noticed that the charm of this arrangement is that the whole figure contains a double set of quartettes, on which the pips are similar.

Solution

[I-79]

This again shows how the stones can be placed so that an ordinary set of dominoes seems to be unduly rich in doublets.

We know how, by the addition of a single letter, our cares can be softened into a caress; but in the following Enigma a still more contradictory result follows, without the addition or alteration of a letter, by a mere separation of syllables:—

Solution

[I-80]

Here are two more perfect arrangements of a set of dominoes in quartettes, so that the pips and blanks are similarly grouped and repeated:—

Solution

[I-81]

Arrange twelve dominoes as is shown in this diagram, and start counting in French from the double five, thus u, n, un; remove the stone you thus reach, which has one pip upon it, and start afresh with the next stone, d, e, u, x, deux; this brings you to the stone with two pips; then t, r, o, i, s, trois, brings you to that with three, and so on until douze brings you to twelve.

Always remove the stone as you hit upon each consecutive number.

Now who can re-arrange these same stones so that a similar result works out in English, thus—o, n, e, one (remove the stone), t, w, o, two, and so on throughout?

Solution

Solution

[I-82]

In this diagram the word EACH is formed by the use of a complete set of stones, placing every letter in proper domino sequence.

There are also the same number of pips in each letter. Can you construct another English word under the same conditions? As a hint, the word that we have in mind is plural.

Solution

Solution

This amusing Charade is from the pen of a wise and witty Irish Bishop:—

Solution

[I-83]

This very beautiful specimen of a knight’s tour on the chess-board takes its name from the figure formed by the tracery at its centre.

An endless number of symmetrical patterns of varied design can be formed, by a knight’s consecutive moves, with patience and ingenuity.

[I-84]

Here is a pretty and very regular specimen of a knight’s tour on the chess board.

It is one of many variations which produce in the tracery a central star.

Solution

[I-85]

Here is a remarkably symmetrical specimen of a knight’s tour on the chess board.

It takes its name from the central archway, which this arrangement forms.

Solution

Solution

[I-86]

In No. LXXXIV we gave a pretty illustration of a knight’s tour, with a central star.

Here is a good course which shows in its symmetrical tracery a pair of stars.

[I-87]

Among the countless fanciful variations of the knight’s tour that are possible, some have been so designed that more than a merely symmetrical pattern is involved.

Here is, for example, an excellent suggestion of the sails of a windmill with their central fittings.

Solution

[I-88]

Here is a very distinctive specimen of the knight’s tour, in which the design reminds us of the old-fashioned lazy-tongs, which stretched out and then back, by opening or shutting their handles on finger and thumb.

Solution

[I-89]

This beautiful symmetrical knight’s tour involves in its accomplishment a pretty problem in arithmetic:—

If we follow the course of the knight step by step, and number consecutively the squares on which it rests at each move, we find that there is a constant difference of 32 between the numbers on any two of these squares that correspond in position on opposite sides of the central line.

Solution

[I-90]

This short symmetrical knight’s tour can be tested on a corner of the chessboard:—

The knight can start from any square, and, taking the course indicated, return on the twentieth move to the starting point.

Solution

Solution