by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
Owner:
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, 23 3 + 24 3 + 25 3 = 204 2 . But it admits the answer, 25 3 + 26 3 + 27 3 + 28 3 + 29 3 = 315 2 . The correct answer, however, requires more heaps, but a smaller aggregate number of blocks. Here it is: 14 3 + 15 3 + ... up to 25 3 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 x 13), can be expressed in two ways, 1,105, (5 x 13 x 17), in four ways, 32,045, (5 * 13 x 17 x 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 p 9 166 factors is subject to another law. We cannot express 25, (5 x 5), in two ways, but only in one; yet 125, (5 x 5 x 5) s can be given in two ways, and so can 625, (5x5x5x5); 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 x 3 x 3 x 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 x 3 x 5, or 9 x 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 130.1 at once saw that this was 2 x 5 x 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.
