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 A binary enigma (Posted on 2011-01-21)
From a pile of 104 tokens two players pick up some tokens, alternating turns and obeying three rules:

1. First player is allowed to take any amount of tokens, but not all of them.
2. Next quantities are limited , allowing the player to take only a divisor of the quantity taken by his partner in the preceding move.
3. The player taking the last token wins.

Devise a winning strategy.

 No Solution Yet Submitted by Ady TZIDON No Rating

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 One way (spoiler) | Comment 1 of 5
Well, clearly the 1st player must leave an even number of stones after the move.  Otherwise, the 2nd player can win by taking 1 stone, after which each player can take only 1 per turn and the 2nd player wins.

Taking 102 and leaving 2 doesn't work.  The 2nd player can take them both.

Taking 100 and leaving 4 doesn't work.  The 2nd player can take them all.

Taking 98 and leaving 6 doesn't work.  The 2nd player can take 2, leaving 4, and the first player will lose whether he takes 1 or 2.

Taking 96 and leaving 8 doesn't work.  The 2nd player can take them all.

Taking 94 and leaving 10 does not work.  The 2nd player can take 2, leaving 8, and he then wins by doing whatever the 1st player does.

Taking 92 and leaving 12 does not work.  The 2nd player can take 4, leaving 8, and he then wins by doing whatever the 1st player does.

Taking 90 and leaving 14 does not work.  The 2nd player can take 2, leaving 12, and he then wins by doing whatever the 1st player does.

Taking 88 and leaving 16 seems to work.  The 2nd player can take 1,2,4,8 or 11.  1 and 11 leave no more choices, and 1 wins.
Otherwise, the 1st player wins by repeating the previous move of the 2nd player.

So, I having not figured out the general rule, but the first player can win by taking 88, and thereafter taking the maximum that he is allowed to take.

There may be other initial winning moves also, but we were only asked to find one.

Edited on January 21, 2011, 4:17 pm
 Posted by Steve Herman on 2011-01-21 16:16:17

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