# Solution

(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.

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### References

#### Project Gutenberg

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

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