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The Putnam Mathematical Competition’s Unsolved Problem

2 pointsby alexyesalmost 11 years ago

1 comment

chriscoolalmost 11 years ago
I think it starts like this:<p>1) at most nb of stone increase by one each time one player plays<p>2) nb of stone cannot decrease unless the board is full of stones<p>3) no one can lose unless the board has been full of stone at least once; this is because until the board has been full of stones it is possible to increase the nb of stone by adding one and this is a new position because of 2)<p>4) as soon as there are n - 1 stones, the player who puts the last stone wins; because there are n - 1 different positions left with n - 1 stones and n - 1 is even
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