(related to Problem: 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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