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?
(In reply to Solution
by David Shin)
I can see that, if the second player gets to play his last card, he will win, since the total sum is n(2n+1) a multiple of (2n+1).
But, why is it that only k choices will lead to a win by the first player?
Posted by e.g.
on 2005-01-10 18:07:22