(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.

Thank you to the contributors under CC BY-SA 4.0!



Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

This eBook is for the use of anyone anywhere in the United States and most other parts of the world at no cost and with almost no restrictions whatsoever. You may copy it, give it away or re-use it under the terms of the Project Gutenberg License included with this edition or online at If you are not located in the United States, you'll have to check the laws of the country where you are located before using this ebook.