(related to Problem: Simple Division)

Subtract every number in turn from every other number, and we get $358$ (twice), $716,$ $1,611,$ $1,253,$ and $895.$ Now, we see at a glance that, as $358$ equals $2 \times 179,$ the only number that can divide in every case without a remainder will be $179.$ On trial, we find that this is such a divisor. Therefore, $179$ is the divisor we want, which always leaves a remainder $164$ in the case of the original numbers given.

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.