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 number that can occur.

If I had asked for the smallest aggregate number of coins, the answer

would 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 number of coins. It is a curious coincidence (nothing

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.

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