# Solution

(related to Problem: The Pebble Game)

In the case of fifteen pebbles, the first player wins if he first takes two. Then when he holds an odd number and leaves $1,$ $8,$ or $9$ he wins, and when he holds an even number and leaves $4,$ $5,$ or $12$ he also wins. He can always do one or other of these things until the end of the game, and so defeat his opponent. In the case of thirteen pebbles, the first player must lose if his opponent plays correctly. In fact, the only numbers with which the first player ought to lose are $5$ and multiples of $8$ added to $5,$ such as $13,$ $21,$ $29,$ etc.

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### References

#### Project Gutenberg

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

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