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Thread: Pebble Challenge

  1. #1
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    Cool Pebble Challenge

    There are 100 pebbles on the table. There are two players, A and B, who move alternatively. Player A moves first. The rules of the game are the same for both players: at each move they can remove one, two, three, four of five pebbles. The winner is the player who takes the last pebble. Who is guaranteed to win provided that he plays properly? Convince me why you think this. Same question if the one who takes the last pebble loses.
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  2. #2
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    Re: Pebble Challenge

    This is a variant of the ancient game of NIm. See this link Nim - Wikipedia, the free encyclopedia for the general solution of Nim and in particular, your variant.
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  3. #3
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    Re: Pebble Challenge

    Quote Originally Posted by sadsadsadsa View Post
    There are 100 pebbles on the table. There are two players, A and B, who move alternatively. Player A moves first. The rules of the game are the same for both players: at each move they can remove one, two, three, four of five pebbles.

    I presume you mean "four or five pebbles".

    The winner is the player who takes the last pebble. Who is guaranteed to win provided that he plays properly? Convince me why you think this. Same question if the one who takes the last pebble loses.
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  4. #4
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    Re: Pebble Challenge

    1) Leave a multiple of six pebbles after your go. The person who goes first can always win.

    2) Leave $6n + 1$ pebbles (for some integer $n$) after your go. Again, the first player can always win.
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