Part 22 (1/2)
Then ab + c = denominator, and a - c, b - c, and a - b will be the three nuet 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 di collector may require If the reader would like one, for exa would satisfy him: 99999999/99990001 and 19999/99990001
131--THE SPANISH MISER
There must have been 386 doubloons in one box, 8,450 in another, and 16,514 in the third, because 386 is the smallest nuate number of coins, the ansould 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 nu 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
Here is the answer that fulfils the conditions:-- A = 4 B = 3,364 C = 6,724 D = 2,116 E = 5,476 F = 8,836 G = 9,409 H = 12,769 I = 16,129 Each of these is a square nu 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
The suands 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 anshere Alfonso has one and the same amount Let us take the cases where Alfonso has 6 doubloons, and see hoe may obtain all the 704 different ways indicated above Here are two tables that will serve as keys to all these answers:-- Table I Table II A = 6 A = 6 B = n B = n C = (63 - 5n) + m C = 1 + m D = (128 + 4n) - 4m D = (376 - 16n) - 4m E = 3 + 3m E = (15n - 183) + 3m
In the first table we may substitute for n any whole nuht or any whole number from 1 to (31 + n) inclusive In the second table n may have the value of any whole nuht or any whole nuives (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 et these two answers:-- A = 6 A = 6 B = 5 B = 13 C = 40 C = 1 D = 140 D = 168 E = 9 E = 12 --- --- 200 doubloons 200 doubloons
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 fro the different numbers for the lettersof Alfonso the nuressions, the co 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 proble the first and last terressions 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
In order that a number of sixpences may not be divisible into a number of equal piles, it is necessary that the nu about a prime number, he in; and I will sho he can always do this, whatever the customer may put in the box, and that therefore the banker in 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 belohat the customer put in Thus, banker puts 40, customer, ill say, adds 6, then transfers from the counter 25 (the square of 5), which leaves 71 in all, a priain 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 prireat ht desire the customer to transfer sufficient to raise the contents of the box to a given nu an absurdity, but breaks the rule that neither knohat the other puts in
135--THE STONEMASON'S PROBLEM
The puzzle amounts to this Find the smallest square number that may be expressed as the su barred As more than three heaps were to be supplied, this condition shuts out the otherwise smallest answer, 23 + 24 + 25 = 204 But it admits the answer, 25 + 26 + 27 + 28 + 29 = 315 The correct answer, however, requires ate number of blocks Here it is: 14 + 15 +up to 25 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
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 hatever
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 co 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 tays, 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 tays, but only in one; yet 125, (5 5 5), can be given in tays, 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 priets 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 ork, because these 3's will theet 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 decolance 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 nuave was 130 I at once saw that this was 2 5 13, and consequently that, as 65 can be expressed in tays (64 + 1 and 49 + 16), 130 can also be expressed in tays, 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 nuether, and their sum will in every case be 160,225
137--A STUDY IN THRIFT
Mrs Sandy McAllister will have to save a tre allowance if she is to win that sixth present that her canny husband promised her And the allowance s The probleher than 36 the units of which le, two triangles, and three triangles, using the complete nuular number is such that if we multiply it by 8 and add 1 the result is an odd square number For example, et 9, 25, 49, 81, 121, which are the squares of the odd numbers 3, 5, 7, 9, 11 Therefore in every case where 8x + 1 = a square nuular This point is dealt with in our puzzle, ”The Battle of Hastings” I will now ain hohen the first solution is found, the others may be discovered without any difficulty First of all, here are the figures:-- 8 1 + 1 = 3 8 6 + 1 = 17 8 35 + 1 = 99 8 204 + 1 = 577 8 1189 + 1 = 3363 8 6930 + 1 = 19601 8 40391 + 1 = 114243 The successive pairs of numbers are found in this way:-- (1 3) + (3 1) = 6 (8 1) + (3 3) = 17 (1 17) + (3 6) = 35 (8 6) + (3 17) = 99 (1 99) + (3 35) = 204 (8 35) + (3 99) = 577 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 forles 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 nural halves of 17, 99, and 577 are 8, 49, and 288
All the nules at will The following little diagralance that every square nuulars, and that the side of one triangle will be the sa square, while the other will be just 1 less
[Illustration +-----------+ +---------+ |/| | /| | / | |/ | |/| | /| | / | |/ | |/| /| / | +---------+ +-----------+ ]
Thus a square les, and the suulars will always make a square In nuulars--1, 3, 6, 10, 15, 21, 28--we find that by adding all the consecutive pairs in turn we get the series of square nu three triangles from our numbers is equally direct, and not at all aactual figures, and just stating that every triangular higher than 6 will forles, and readers will know fro 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
+----------------------+-----------+---------------+-----------------------+ | Number | Side of | Side of | Sides of Two | Sides of Three | | | Square | Triangle | Triangles | Triangles | +------------+---------+-----------+---------------+-----------------------+ | 36 | 6 | 8 | 6 + 5 | 5 + 5 + 3 | | 1225 | 35 | 49 | 36 + 34 | 33 + 32 + 16 | | 41616 | 204 | 288 | 204 + 203 | 192 + 192 + 95 | | 1413721 | 1189 | 1681 | 1189 + 1188 | 1121 + 1120 + 560 | | 48024900 | 6930 | 9800 | 6930 + 6929 | 6533 + 6533 + 3267 | | 1631432881 | 40391 | 57121 | 40391 + 40390 | 38081 + 38080 + 19040 | +------------+---------+-----------+---------------+-----------------------+ I should perhaps explain that the arrangeiven in the last two colules 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
We were required to find the sround to form a perfect square, and could pile into a square pyramid I will try to make the matter clear to the merest novice
1 2 3 4 5 6 7 1 3 6 10 15 21 28 1 4 10 20 35 56 84 1 5 14 30 55 91 140 Here in the first roe place in regular order the natural numbers Each number in the second row represents the su to the nuether, make 10 The third row is formed in exactly the same way as the second In the fourth row every nuether the nu number Thus 4 and 10 make 14, 20 and 35 ular numbers, which means that these nuround so as to forles The nuular pyramids, while the numbers in the fourth roill all for the above numbers shows us that every square pyraular 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 for out the series until we come to a square number does not appeal to the mathematical mind, but it serves to sho 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 eneral expression for square pyramid numbers is (2n + 3n + n)/6 For this expression to be also a square number (the special case of 1 excepted) it is necessary that n = p - 1 = 6t, where 2p - 1 = q (the ”Pellian Equation”) In the case of our solution above, n = 24, p = 5, t = 2, q = 7
139--THE DUTCHMEN'S WIVES
The s, 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 nu paid by each woman, and 8, 12, and 32 the same in the case of their respective husbands Froiven as to their purchases, we can now pair theht 8 and 1; Elas and Katrun 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 knoe may determine the maximum number of ways in which a number may be expressed as the difference between two squares, and hoe are to find the actual squares Any integer except 1, 4, and twice any odd nural 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 23511 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 nuht pairs of factors of which are 1 1485, 3 495, 5 297, 9 165, 11 135, 15 99, 27 55, and 33 45 The suive 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 suive 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 nuive 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 inso expressed it, we may at once obtain the factors, as we have seen Fermat showed in a letter to Mersenne or Frenicle, in 1643, hoe may discover whether a number may be expressed as the difference of two squares in more than one way, or proved to be a prie nuh in practice it may be considerably shortened Inlarge numbers, and I have always held the opinion that Fer that is historical and wrapped in irls' names were Ada Smith, Annie Brown, Emily Jones, Mary Robinson, and Bessie Evans
141--SAturdAY MARKETING
As every person's purchase was of the value of an exact nus, and as the party possessed when they started out forty shi+lling coins altogether, there was no necessity for any lady to have any se, 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 shi+llings left, which ht persons in coin without any change being required
142--THE SILK PATCHWORK
[Illustration]
Our illustration will sho 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 kno to asse 13 in the article
[Illustration: Fig 1]
[Illustration: Fig 2]
143--TWO CROSSES FROM ONE
It will be seen that one cross is cut out entire, as A in Fig 1, while the four pieces2, which will be of exactly the same size as the other I will leave the reader the pleasant task of discovering for hi the direction of the cuts Note that the Swastika again appears
The difficult question now presents itself: How are we to cut three Greek crosses from one in the fewest possible pieces? As a matter of fact, this problem may be solved in as few as thirteen pieces; but as I know lad to have so to work on of which they are not shown the solution, I leave the mystery for the present undisclosed