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Who will win? (Posted on 2005-01-10) Difficulty: 4 of 5
There are 2n cards labeled 1, 2, ..., 2n respectively, and the cards are distributed randomly between two players so that each has n cards. Each player takes turns to place one card, and you win if you put down a card so that the current sum of all the played cards is divisible by 2n+1.

For example, if n=10, and the previously placed cards are 5, 8, 9, 19, then if player A now places 1, he wins since 5+8+9+19+1 = 42 is divisible by 2*10+1=21.

Assuming both players want to win, what strategy should one adopt in order to win? Following the strategy, is there a consistent winner of this game?

See The Solution Submitted by Bon    
Rating: 2.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Solution | Comment 4 of 9 |
(In reply to re: Solution by e.g.)

If Player 2 has (k+1) choices, Player 1 has k choices on his next turn.  Suppose that all (k+1) choices by Player 2 lead to a win for Player 1.  Then there are distinct x,y, such that if Player 2 plays either, then Player 1 can respond with z to win.  This means that x+z = y+z (mod 2n+1), or x=y (mod 2n+1), contradicting the fact that x and y are distinct.

Edited on January 10, 2005, 6:55 pm
  Posted by David Shin on 2005-01-10 18:52:43

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