an infinite number of different answers, but on consideration I found
that this was not the case. I discovered that while every box contained
coins, the contents of A, B, C increased in weight in alphabetical
order; so did D, E, F; and so did G, H, I; but D or E need not be
heavier than C, nor G or H heavier than F. It was also perfectly certain
that box A could not contain more than a dozen coins at the outside;
there might not be half that number, but I was positive that there were
not more than twelve. With this knowledge I was able to arrive at the
correct answer.

In short, we have to discover nine square numbers such that A, B, C; and
D, E, F; and G, H, I are three groups in arithmetical progression, the
common difference being the same in each group, and A being less than
12. How many doubloons were there in every one of the nine boxes?


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 × 12 = 72