distri-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:—
| Table I. | Table II. |
| A=6. | A=6. |
| B=n. | B=n |
| C=(63-5n)+m. | C=1+m. |
| D=(118+4n)—4m. | D=(376-16n)-4m. |
| E=3+3m. | E=(15n-183)+3m. |
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:—
| 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 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, on 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 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.
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 I 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.
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 he 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
