(related to Problem: The Bag of Nuts)

It will be found that when Herbert takes twelve, Robert and Christopher will take nine and fourteen respectively and that they will have together taken thirty-five nuts. As $35$ is contained in $770$ twenty-two times, we have merely to multiply $12, 9,$ and $14$ by $22$ to discover that Herbert's share was $264,$ Robert's $198,$ and Christopher's $308.$ Then, as the total of their ages is $17\frac 12$ years or half the sum of $12, 9,$ and $14,$ their respective ages must be $6,$ $4\frac 12,$ and $7$ years.

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.