The 24-stone game where taking the last stone loses

Question

A pile holds 24 stones. Two players alternate; on your turn you must remove 1, 2, or 3 stones. This time there is a twist: the player who is forced to take the last stone loses. If you move first, can you force a win, and how?

Accepted Answer

Yes, the first player wins. The key is to leave your opponent facing a pile whose size is 1 more than a multiple of 4 (that is, \equiv 1 \pmod 4: sizes 1, 5, 9, 13, 17, 21). Why those are the losing positions. Look at the smallest pile first. A pile of 1 is a disaster for whoever must move, they have to take that single last stone and lose. Now consider 5: whatever you take (1, 2, or 3), you leave 4, 3, or 2, and from any of those your opponent can hand exactly 1 back to you. So being forced to move at 5 also loses. The same reasoning marches up in steps of 4: from a pile \equiv 1 \pmod 4, ev…

The 24-stone game where taking the last stone loses

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