133.—­THE FIVE BRIGANDS.

The five Spanish brigands, Alfonso, Benito, Carlos, Diego, and Esteban, were counting their spoils after a raid, when it was found that they had captured altogether exactly 200 doubloons.  One of the band pointed out that if Alfonso had twelve times as much, Benito three times as much, Carlos the same amount, Diego half as much, and Esteban one-third as much, they would still have altogether just 200 doubloons.  How many doubloons had each?

There are a good many equally correct answers to this question.  Here is one of them: 

    A 6 x 12 = 72
    B 12 x 3 = 36
    C 17 x 1 = 17
    D 120 x 1/2 = 60
    E 45 x 1/3 = 15
       ___ ___
       200 200

The puzzle is to discover exactly how many different answers there are, it being understood that every man had something and that there is to be no fractional money—­only doubloons in every case.

This problem, worded somewhat differently, was propounded by Tartaglia (died 1559), and he flattered himself that he had found one solution; but a French mathematician of note (M.A.  Labosne), in a recent work, says that his readers will be astonished when he assures them that there are 6,639 different correct answers to the question.  Is this so?  How many answers are there?

134.—­THE BANKER’S PUZZLE.

A banker had a sporting customer who was always anxious to wager on anything.  Hoping to cure him of his bad habit, he proposed as a wager that the customer would not be able to divide up the contents of a box containing only sixpences into an exact number of equal piles of sixpences.  The banker was first to put in one or more sixpences (as many as he liked); then the customer was to put in one or more (but in his case not more than a pound in value), neither knowing what the other put in.  Lastly, the customer was to transfer from the banker’s counter to the box as many sixpences as the banker desired him to put in.  The puzzle is to find how many sixpences the banker should first put in and how many he should ask the customer to transfer, so that he may have the best chance of winning.

135.—­THE STONEMASON’S PROBLEM.

A stonemason once had a large number of cubic blocks of stone in his yard, all of exactly the same size.  He had some very fanciful little ways, and one of his queer notions was to keep these blocks piled in cubical heaps, no two heaps containing the same number of blocks.  He had discovered for himself (a fact that is well known to mathematicians) that if he took all the blocks contained in any number of heaps in regular order, beginning with the single cube, he could always arrange those on the ground so as to form a perfect square.  This will be clear to the reader, because one block is a square, 1 + 8 = 9 is a square, 1 + 8 + 27 = 36 is a square, 1 + 8 + 27 + 64 = 100 is a square, and so on.  In fact, the sum of any number of consecutive cubes, beginning always with 1, is in every case a square number.