(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.
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 http://www.gutenberg.org. 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.
