The lowest possible amount, and therefore the correct answer, is £19,345, distributed in an ordinary year that began on a Sunday. The last year of this kind was 1911. He would have paid £53 on every day of the year, or £62 on every weekday, with £1 left over, as required, in the latter event.

117.—A FENCE PROBLEM.—solution

Though this puzzle presents no great difficulty to any one possessing a knowledge of algebra, it has perhaps rather interesting features.

Seeing, as one does in the illustration, just one corner of the proposed square, one is scarcely prepared for the fact that the field, in order to comply with the conditions, must contain exactly 501,760 acres, the fence requiring the same number of rails. Yet this is the correct answer, and the only answer, and if that gentleman in Iowa carries out his intention, his field will be twenty-eight miles long on each side, and a little larger than the county of Westmorland. I am not aware that any limit has ever been fixed to the size of a "field," though they do not run so large as this in Great Britain. Still, out in Iowa, where my correspondent resides, they do these things on a very big scale. I have, however, reason to believe that when he finds the sort of task he has set himself, he will decide to abandon it; for if that cow decides to roam to fresh woods and pastures new, the milkmaid may have to start out a week in advance in order to obtain the morning's milk.

Here is a little rule that will always apply where the length of the rail is half a pole. Multiply the number of rails in a hurdle by four, and the result is the exact number of miles in the side of a square field containing the same number of acres as there are rails in the complete fence. Thus, with a one-rail fence the field is four miles square; a two-rail fence gives eight miles square; a three-rail fence, twelve miles square; and so on, until we find that a seven-rail fence multiplied by four gives a field of twenty-eight miles square. In the case of our present problem, if the field be made smaller, then the number of rails will exceed the number of acres; while if the field be made larger, the number of rails will be less than the acres of the field.

118.—CIRCLING THE SQUARES.—solution

Though this problem might strike the novice as being rather difficult, it is, as a matter of fact, quite easy, and is made still easier by inserting four out of the ten numbers.

First, it will be found that squares that are diametrically opposite have a common difference. For example, the difference between the square of 14 and the square of 2, in the diagram, is 192; and the difference between the square of 16 and the square of 8 is also 192. This must be so in every case. Then it should be remembered that the difference between squares of two consecutive numbers is always twice the smaller number plus 1, and that the difference between the squares of any two numbers can always be expressed as the difference of the numbers multiplied by their sum. Thus the square of 5 (25) less the square of 4 (16) equals (2 × 4) + 1, or 9; also, the square of 7 (49) less the square of 3 (9) equals (7 + 3) × (7 - 3), or 40.

Now, the number 192, referred to above, may be divided into five different pairs of even factors: 2 × 96, 4 × 48, 6 × 32, 8 × 24, and 12 × 16, and these divided by 2 give us, 1 × 48, 2 × 24, 3 × 16, 4 × 12, and 6 × 8. The difference and sum respectively of each of these pairs in turn produce 47, 49; 22, 26; 13, 19; 8, 16; and 2, 14. These are the required numbers, four of which are already placed. The six numbers that have to be added may be placed in just six different ways, one of which is as follows, reading round the circle clockwise: 16, 2, 49, 22, 19, 8, 14, 47, 26, 13.

I will just draw the reader's attention to one other little point. In all circles of this kind, the difference between diametrically opposite numbers increases by a certain ratio, the first numbers (with the exception of a circle of 6) being 4 and 6, and the others formed by doubling the next preceding but one. Thus, in the above case, the first difference is 2, and then the numbers increase by 4, 6, 8, and 12. Of course, an infinite number of solutions may be found if we admit fractions. The number of squares in a circle of this kind must, however, be of the form 4n + 6; that is, it must be a number composed of 6 plus a multiple of 4.

119.—RACKBRANE'S LITTLE LOSS.—solution

The professor must have started the game with thirteen shillings, Mr. Potts with four shillings, and Mrs. Potts with seven shillings.

120.—THE FARMER AND HIS SHEEP.—solution

The farmer had one sheep only! If he divided this sheep (which is best done by weight) into two parts, making one part two-thirds and the other part one-third, then the difference between these two numbers is the same as the difference between their squares—that is, one-third. Any two fractions will do if the denominator equals the sum of the two numerators.

121.—HEADS OR TAILS.—solution

Crooks must have lost, and the longer he went on the more he would lose. In two tosses he would be left with three-quarters of his money, in four tosses with nine-sixteenths of his money, in six tosses with twenty-seven sixty-fourths of his money, and so on. The order of the wins and losses makes no difference, so long as their number is in the end equal.

122.—THE SEE-SAW PUZZLE.—solution

The boy's weight must have been about 39.79 lbs. A brick weighed 3 lbs. Therefore 16 bricks weighed 48 lbs. and 11 bricks 33 lbs. Multiply 48 by 33 and take the square root.

123.—A LEGAL DIFFICULTY.—solution

It was clearly the intention of the deceased to give the son twice as much as the mother, or the daughter half as much as the mother. Therefore the most equitable division would be that the mother should take two-sevenths, the son four-sevenths, and the daughter one-seventh.

124.—A QUESTION OF DEFINITION.—solution

There is, of course, no difference in area between a mile square and a square mile. But there may be considerable difference in shape. A mile square can be no other shape than square; the expression describes a surface of a certain specific size and shape. A square mile may be of any shape; the expression names a unit of area, but does not prescribe any particular shape.

125.—THE MINERS' HOLIDAY.—solution

Bill Harris must have spent thirteen shillings and sixpence, which would be three shillings more than the average for the seven men—half a guinea.

126.—SIMPLE MULTIPLICATION.—solution

The number required is 3,529,411,764,705,882, which may be multiplied by 3 and divided by 2, by the simple expedient of removing the 3 from one end of the row to the other. If you want a longer number, you can increase this one to any extent by repeating the sixteen figures in the same order.

127.—SIMPLE DIVISION.—solution

Subtract every number in turn from every other number, and we get 358 (twice), 716, 1,611, 1,253, and 895. Now, we see at a glance that, as 358 equals 2 × 179, the only number that can divide in every case without a remainder will be 179. On trial we find that this is such a divisor. Therefore, 179 is the divisor we want, which always leaves a remainder 164 in the case of the original numbers given.

128.—A PROBLEM IN SQUARES.—solution

The sides of the three boards measure 31 in., 41 in., and 49 in. The common difference of area is exactly five square feet. Three numbers whose squares are in A.P., with a common difference of 7, are ¹¹³/₁₂₀, ³³⁷/₁₂₀, ⁴⁶³/₁₂₀; and with a common difference of 13 are ⁸⁰⁹²⁹/₁₉₃₈₀, ¹⁰⁶⁹²¹/₁₉₃₈₀, and ¹²⁷⁷²⁹/₁₉₃₈₀. In the case of whole square numbers the common difference will always be divisible by 24, so it is obvious that our squares must be fractional. Readers should now try to solve the case where the common difference is 23. It is rather a hard nut.

129.—THE BATTLE OF HASTINGS.—solution

Any number (not itself a square number) may be multiplied by a square that will give a product 1 less than another square. The given number must not itself be a square, because a square multiplied by a square produces a square, and no square plus 1 can be a square. My remarks throughout must be understood to apply to whole numbers, because fractional soldiers are not of much use in war.

Now, of all the numbers from 2 to 99 inclusive, 61 happens to be the most awkward one to work, and the lowest possible answer to our puzzle is that Harold's army consisted of 3,119,882,982,860,264,400 men. That is, there would be 51,145,622,669,840,400 men (the square of 226,153,980) in each of the sixty-one squares. Add one man (Harold), and they could then form one large square with 1,766,319,049 men on every side. The general problem, of which this is a particular case, is known as the "Pellian Equation"—apparently because Pell neither first propounded the question nor first solved it! It was issued as a challenge by Fermat to the English mathematicians of his day. It is readily solved by the use of continued fractions.

Next to 61, the most difficult number under 100 is 97, where 97 × 6,377,352² + 1 = a square.

The reason why I assumed that there must be something wrong with the figures in the chronicle is that we can confidently say that Harold's army did not contain over three trillion men! If this army (not to mention the Normans) had had the whole surface of the earth (sea included) on which to encamp, each man would have had slightly more than a quarter of a square inch of space in which to move about! Put another way: Allowing one square foot of standing-room per man, each small square would have required all the space allowed by a globe three times the diameter of the earth.

130.—THE SCULPTOR'S PROBLEM.—solution

A little thought will make it clear that the answer must be fractional, and that in one case the numerator will be greater and in the other case less than the denominator. As a matter of fact, the height of the larger cube must be ⁸/₇ ft., and of the smaller ³/₇ ft., if we are to have the answer in the smallest possible figures. Here the lineal measurement is ¹¹/₇ ft.—that is, 1⁴/₇ ft. What are the cubic contents of the two cubes? First ⁸/₇ × ³/₇ × ⁸/₇ = ⁵¹²/₃₄₃, and secondly ³/₇ × ³/₇ × ³/₇ = ²⁷/₃₄₃. Add these together and the result is ⁵³⁹/₃₄₃, which reduces to ¹¹/₇ or 1⁴/₇ ft. We thus see that the answers in cubic feet and lineal feet are precisely the same.

The germ of the idea is to be found in the works of Diophantus of Alexandria, who wrote about the beginning of the fourth century. These fractional numbers appear in triads, and are obtained from three generators, a, b, c, where a is the largest and c the smallest.

Then ab+c²=denominator, and a²-c², b²-c², and a²-b² will be the three numerators. Thus, using the generators 3, 2, 1, we get ⁸/₇, ³/₇, ⁵/₇ and we can pair the first and second, as in the above solution, or the first and third for a second solution. The denominator must always be a prime number of the form 6n+1, or composed of such primes. Thus you can have 13, 19, etc., as denominators, but not 25, 55, 187, etc.

When the principle is understood there is no difficulty in writing down the dimensions of as many sets of cubes as the most exacting collector may require. If the reader would like one, for example, with plenty of nines, perhaps the following would satisfy him: ^99999999/_99990001 and ¹⁹⁹⁹⁹/_99990001.

131.—THE SPANISH MISER.—solution

There must have been 386 doubloons in one box, 8,450 in another, and 16,514 in the third, because 386 is the smallest number that can occur. If I had asked for the smallest aggregate number of coins, the answer would have been 482, 3,362, and 6,242. It will be found in either case that if the contents of any two of the three boxes be combined, they form a square number of coins. It is a curious coincidence (nothing more, for it will not always happen) that in the first solution the digits of the three numbers add to 17 in every case, and in the second solution to 14. It should be noted that the middle one of the three numbers will always be half a square.

132.—THE NINE TREASURE BOXES.—solution

Here is the answer that fulfils the conditions:—

Each of these is a square number, the roots, taken in alphabetical order, being 2, 58, 82, 46, 74, 94, 97, 113, and 127, while the required difference between A and B, B and C, D and E. etc., is in every case 3,360.

133.—THE FIVE BRIGANDS.—solution

The sum of 200 doubloons might have been held by the five brigands in any one of 6,627 different ways. Alfonso may have held any number from 1 to 11. If he held 1 doubloon, there are 1,005 different ways of distributing the remainder; if he held 2, there are 985 ways; if 3, there are 977 ways; if 4, there are 903 ways; if 5 doubloons, 832 ways; if 6 doubloons, 704 ways; if 7 doubloons, 570 ways; if 8 doubloons, 388 ways; if 9 doubloons, 200 ways; if 10 doubloons, 60 ways; and if Alfonso held 11 doubloons, the remainder could be distributed in 3 different ways. More than 11 doubloons he could not possibly have had. It will scarcely be expected that I shall give all these 6,627 ways at length. What I propose to do is to enable the reader, if he should feel so disposed, to write out all the answers where Alfonso has one and the same amount. Let us take the cases where Alfonso has 6 doubloons, and see how we may obtain all the 704 different ways indicated above. Here are two tables that will serve as keys to all these answers:—

In the first table we may substitute for n any whole number from 1 to 12 inclusive, and m may be nought or any whole number from 1 to (31 + n) inclusive. In the second table n may have the value of any whole number from 13 to 23 inclusive, and m may be nought or any whole number from 1 to (93 - 4n) inclusive. The first table thus gives (32 + n) answers for every value of n; and the second table gives (94 - 4n) answers for every value of n. The former, therefore, produces 462 and the latter 242 answers, which together make 704, as already stated.

Let us take Table I., and say n = 5 and m = 2; also in Table II. take n = 13 and m = 0. Then we at once get these two answers:—

These will be found to work correctly. All the rest of the 704 answers, where Alfonso always holds six doubloons, may be obtained in this way from the two tables by substituting the different numbers for the letters m and n.

Put in another way, for every holding of Alfonso the number of answers is the sum of two arithmetical progressions, the common difference in one case being 1 and in the other -4. Thus in the case where Alfonso holds 6 doubloons one progression is 33 + 34 + 35 + 36 + ... + 43 + 44, and the other 42 + 38 + 34 + 30 + ... + 6 + 2. The sum of the first series is 462, and of the second 242—results which again agree with the figures already given. The problem may be said to consist in finding the first and last terms of these progressions. I should remark that where Alfonso holds 9, 10, or 11 there is only one progression, of the second form.

134.—THE BANKER'S PUZZLE.—solution

In order that a number of sixpences may not be divisible into a number of equal piles, it is necessary that the number should be a prime. If the banker can bring about a prime number, he will win; and I will show how he can always do this, whatever the customer may put in the box, and that therefore the banker will win to a certainty. The banker must first deposit forty sixpences, and then, no matter how many the customer may add, he will desire the latter to transfer from the counter the square of the number next below what the customer put in. Thus, banker puts 40, customer, we will say, adds 6, then transfers from the counter 25 (the square of 5), which leaves 71 in all, a prime number. Try again. Banker puts 40, customer adds 12, then transfers 121 (the square of 11), as desired, which leaves 173, a prime number. The key to the puzzle is the curious fact that any number up to 39, if added to its square and the sum increased by 41, makes a prime number. This was first discovered by Euler, the great mathematician. It has been suggested that the banker might desire the customer to transfer sufficient to raise the contents of the box to a given number; but this would not only make the thing an absurdity, but breaks the rule that neither knows what the other puts in.

135.—THE STONEMASON'S PROBLEM.—solution

The puzzle amounts to this. Find the smallest square number that may be expressed as the sum of more than three consecutive cubes, the cube 1 being barred. As more than three heaps were to be supplied, this condition shuts out the otherwise smallest answer, 23³ + 24³ + 25³ = 204². But it admits the answer, 25³ + 26³ + 27³ + 28³ + 29³ = 315². The correct answer, however, requires more heaps, but a smaller aggregate number of blocks. Here it is: 14³ + 15³ + ... up to 25³ inclusive, or twelve heaps in all, which, added together, make 97,344 blocks of stone that may be laid out to form a square 312 × 312. I will just remark that one key to the solution lies in what are called triangular numbers. (See pp. 13, 25, and 166.)

136.—THE SULTAN'S ARMY.—solution

The smallest primes of the form 4n + 1 are 5, 13, 17, 29, and 37, and the smallest of the form 4n - 1 are 3, 7, 11, 19, and 23. Now, primes of the first form can always be expressed as the sum of two squares, and in only one way. Thus, 5 = 4 + 1; 13 = 9 + 4; 17 = 16 + 1; 29 = 25 + 4; 37 = 36 + 1. But primes of the second form can never be expressed as the sum of two squares in any way whatever.

In order that a number may be expressed as the sum of two squares in several different ways, it is necessary that it shall be a composite number containing a certain number of primes of our first form. Thus, 5 or 13 alone can only be so expressed in one way; but 65, (5 × 13), can be expressed in two ways, 1,105, (5 × 13 × 17), in four ways, 32,045, (5 × 13 × 17 × 29), in eight ways. We thus get double as many ways for every new factor of this form that we introduce. Note, however, that I say new factor, for the repetition of factors is subject to another law. We cannot express 25, (5 × 5), in two ways, but only in one; yet 125, (5 × 5 × 5), can be given in two ways, and so can 625, (5 × 5 × 5 × 5); while if we take in yet another 5 we can express the number as the sum of two squares in three different ways.

If a prime of the second form gets into your composite number, then that number cannot be the sum of two squares. Thus 15, (3 × 5), will not work, nor will 135, (3 × 3 × 3 × 5); but if we take in an even number of 3's it will work, because these 3's will themselves form a square number, but you will only get one solution. Thus, 45, (3 × 3 × 5, or 9 × 5) = 36 + 9. Similarly, the factor 2 may always occur, or any power of 2, such as 4, 8, 16, 32; but its introduction or omission will never affect the number of your solutions, except in such a case as 50, where it doubles a square and therefore gives you the two answers, 49 + 1 and 25 + 25.

Now, directly a number is decomposed into its prime factors, it is possible to tell at a glance whether or not it can be split into two squares; and if it can be, the process of discovery in how many ways is so simple that it can be done in the head without any effort. The number I gave was 130. I at once saw that this was 2 × 5 × 13, and consequently that, as 65 can be expressed in two ways (64 + 1 and 49 + 16), 130 can also be expressed in two ways, the factor 2 not affecting the question.

The smallest number that can be expressed as the sum of two squares in twelve different ways is 160,225, and this is therefore the smallest army that would answer the Sultan's purpose. The number is composed of the factors 5 × 5 × 13 × 17 × 29, each of which is of the required form. If they were all different factors, there would be sixteen ways; but as one of the factors is repeated, there are just twelve ways. Here are the sides of the twelve pairs of squares: (400 and 15), (399 and 32), (393 and 76), (392 and 81), (384 and 113), (375 and 140), (360 and 175), (356 and 183), (337 and 216), (329 and 228), (311 and 252), (265 and 300). Square the two numbers in each pair, add them together, and their sum will in every case be 160,225.

137.—A STUDY IN THRIFT.—solution

Mrs. Sandy McAllister will have to save a tremendous sum out of her housekeeping allowance if she is to win that sixth present that her canny husband promised her. And the allowance must be a very liberal one if it is to admit of such savings. The problem required that we should find five numbers higher than 36 the units of which may be displayed so as to form a square, a triangle, two triangles, and three triangles, using the complete number in every one of the four cases.

Every triangular number is such that if we multiply it by 8 and add 1 the result is an odd square number. For example, multiply 1, 3, 6, 10, 15 respectively by 8 and add 1, and we get 9, 25, 49, 81, 121, which are the squares of the odd numbers 3, 5, 7, 9, 11. Therefore in every case where 8x² + 1 = a square number, x² is also a triangular. This point is dealt with in our puzzle, "The Battle of Hastings." I will now merely show again how, when the first solution is found, the others may be discovered without any difficulty. First of all, here are the figures:—

The successive pairs of numbers are found in this way:—

and so on. Look for the numbers in the table above, and the method will explain itself.

Thus we find that the numbers 36, 1225, 41616, 1413721, 48024900, and 1631432881 will form squares with sides of 6, 35, 204, 1189, 6930, and 40391; and they will also form single triangles with sides of 8, 49, 288, 1681, 9800, and 57121. These numbers may be obtained from the last column in the first table above in this way: simply divide the numbers by 2 and reject the remainder. Thus the integral halves of 17, 99, and 577 are 8, 49, and 288.

All the numbers we have found will form either two or three triangles at will. The following little diagram will show you graphically at a glance that every square number must necessarily be the sum of two triangulars, and that the side of one triangle will be the same as the side of the corresponding square, while the other will be just 1 less.

Thus a square may always be divided easily into two triangles, and the sum of two consecutive triangulars will always make a square. In numbers it is equally clear, for if we examine the first triangulars—1, 3, 6, 10, 15, 21, 28—we find that by adding all the consecutive pairs in turn we get the series of square numbers—9, 16, 25, 36, 49, etc.

The method of forming three triangles from our numbers is equally direct, and not at all a matter of trial. But I must content myself with giving actual figures, and just stating that every triangular higher than 6 will form three triangulars. I give the sides of the triangles, and readers will know from my remarks when stating the puzzle how to find from these sides the number of counters or coins in each, and so check the results if they so wish.

I should perhaps explain that the arrangements given in the last two columns are not the only ways of forming two and three triangles. There are others, but one set of figures will fully serve our purpose. We thus see that before Mrs. McAllister can claim her sixth £5 present she must save the respectable sum of £1,631,432,881.

138.—THE ARTILLERYMEN'S DILEMMA.—solution

We were required to find the smallest number of cannon balls that we could lay on the ground to form a perfect square, and could pile into a square pyramid. I will try to make the matter clear to the merest novice.

Here in the first row we place in regular order the natural numbers. Each number in the second row represents the sum of the numbers in the row above, from the beginning to the number just over it. Thus 1, 2, 3, 4, added together, make 10. The third row is formed in exactly the same way as the second. In the fourth row every number is formed by adding together the number just above it and the preceding number. Thus 4 and 10 make 14, 20 and 35 make 55. Now, all the numbers in the second row are triangular numbers, which means that these numbers of cannon balls may be laid out on the ground so as to form equilateral triangles. The numbers in the third row will all form our triangular pyramids, while the numbers in the fourth row will all form square pyramids.

Thus the very process of forming the above numbers shows us that every square pyramid is the sum of two triangular pyramids, one of which has the same number of balls in the side at the base, and the other one ball fewer. If we continue the above table to twenty-four places, we shall reach the number 4,900 in the fourth row. As this number is the square of 70, we can lay out the balls in a square, and can form a square pyramid with them. This manner of writing out the series until we come to a square number does not appeal to the mathematical mind, but it serves to show how the answer to the particular puzzle may be easily arrived at by anybody. As a matter of fact, I confess my failure to discover any number other than 4,900 that fulfils the conditions, nor have I found any rigid proof that this is the only answer. The problem is a difficult one, and the second answer, if it exists (which I do not believe), certainly runs into big figures.

For the benefit of more advanced mathematicians I will add that the general expression for square pyramid numbers is (2n³ + 3n² + n)/6. For this expression to be also a square number (the special case of 1 excepted) it is necessary that n = p² - 1 = 6t², where 2p² - 1 = q² (the "Pellian Equation"). In the case of our solution above, n = 24, p = 5, t = 2, q = 7.

139.—THE DUTCHMEN'S WIVES.—solution

The money paid in every case was a square number of shillings, because they bought 1 at 1s., 2 at 2s., 3 at 3s., and so on. But every husband pays altogether 63s. more than his wife, so we have to find in how many ways 63 may be the difference between two square numbers. These are the three only possible ways: the square of 8 less the square of 1, the square of 12 less the square of 9, and the square of 32 less the square of 31. Here 1, 9, and 31 represent the number of pigs bought and the number of shillings per pig paid by each woman, and 8, 12, and 32 the same in the case of their respective husbands. From the further information given as to their purchases, we can now pair them off as follows: Cornelius and Gurtrün bought 8 and 1; Elas and Katrün bought 12 and 9; Hendrick and Anna bought 32 and 31. And these pairs represent correctly the three married couples.

The reader may here desire to know how we may determine the maximum number of ways in which a number may be expressed as the difference between two squares, and how we are to find the actual squares. Any integer except 1, 4, and twice any odd number, may be expressed as the difference of two integral squares in as many ways as it can be split up into pairs of factors, counting 1 as a factor. Suppose the number to be 5,940. The factors are 2².3³.5.11. Here the exponents are 2, 3, 1, 1. Always deduct 1 from the exponents of 2 and add 1 to all the other exponents; then we get 1, 4, 2, 2, and half the product of these four numbers will be the required number of ways in which 5,940 may be the difference of two squares—that is, 8. To find these eight squares, as it is an even number, we first divide by 4 and get 1485, the eight pairs of factors of which are 1 × 1485, 3 × 495, 5 × 297, 9 × 165, 11 × 135, 15 × 99, 27 × 55, and 33 × 45. The sum and difference of any one of these pairs will give the required numbers. Thus, the square of 1,486 less the square of 1,484 is 5,940, the square of 498 less the square of 492 is the same, and so on. In the case of 63 above, the number is odd; so we factorize at once, 1 × 63, 3 × 21, 7 × 9. Then we find that half the sum and difference will give us the numbers 32 and 31, 12 and 9, and 8 and 1, as shown in the solution to the puzzle.

The reverse problem, to find the factors of a number when you have expressed it as the difference of two squares, is obvious. For example, the sum and difference of any pair of numbers in the last sentence will give us the factors of 63. Every prime number (except 1 and 2) may be expressed as the difference of two squares in one way, and in one way only. If a number can be expressed as the difference of two squares in more than one way, it is composite; and having so expressed it, we may at once obtain the factors, as we have seen. Fermat showed in a letter to Mersenne or Frénicle, in 1643, how we may discover whether a number may be expressed as the difference of two squares in more than one way, or proved to be a prime. But the method, when dealing with large numbers, is necessarily tedious, though in practice it may be considerably shortened. In many cases it is the shortest method known for factorizing large numbers, and I have always held the opinion that Fermat used it in performing a certain feat in factorizing that is historical and wrapped in mystery.

140.—FIND ADA'S SURNAME.—solution

The girls' names were Ada Smith, Annie Brown, Emily Jones, Mary Robinson, and Bessie Evans.

141.—SATURDAY MARKETING.—solution

As every person's purchase was of the value of an exact number of shillings, and as the party possessed when they started out forty shilling coins altogether, there was no necessity for any lady to have any smaller change, or any evidence that they actually had such change. This being so, the only answer possible is that the women were named respectively Anne Jones, Mary Robinson, Jane Smith, and Kate Brown. It will now be found that there would be exactly eight shillings left, which may be divided equally among the eight persons in coin without any change being required.

142.—THE SILK PATCHWORK.—solution

Our illustration will show how to cut the stitches of the patchwork so as to get the square F entire, and four equal pieces, G, H, I, K, that will form a perfect Greek cross. The reader will know how to assemble these four pieces from Fig. 13 in the article.

143.—TWO CROSSES FROM ONE.—solution

It will be seen that one cross is cut out entire, as A in Fig. 1, while the four pieces marked B, C, D and E form the second cross, as in Fig. 2, which will be of exactly the same size as the other. I will leave the reader the pleasant task of discovering for himself the best way of finding the direction of the cuts. Note that the Swastika again appears.