(related to Problem: The Pierrot's Puzzle)

There are just six different solutions to this puzzle, as follows: — * $8\times 473=3784$ * $9\times 351=3159$ * $15\times 93=1395$ * $21\times 87=1287$ * $27\times 81=2187$ * $35\times 41=1435$

It will be seen that in every case the two multipliers contain exactly the same figures as the product.

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Project Gutenberg

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

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