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Boxes and Coins Game?
There is a game involving 5 boxes numbered 1, 2, 3, 4, and 5 containing 0, 20, 30, 40, and 50 coins respectively. The game has two players (A and B) who take turns alternately. It begins with A taking some coins from some box n and moving them to box n-1. (For example A could take 23 coins from box 4 and move them to box 3 so that the boxes have 0, 20, 53, 17, 50 coins respectively. Then B might take 2 coins from box 2 and move them to box 1 so that the boxes have 2, 18, 53, 17, 50 coins respectively.) You cannot move any coins that are in box 1. When all the coins have been moved to box 1 the game is over. The player who makes the last move wins. Which player has a winning strategy and what is it?
Hint: Look at much smaller cases first such as (a) 0 0 1 0 0, 0 0 2 0 0, 0 0 3 0 0, etc, and (b) 0 1 0 1 0, 0 2 0 2 0, 0 3 0 3 0, etc.
I think it's about time to close this one. Potter Boy does have the correct answer that A has a winning strategy.
I'll offer the strategy as well. Player A can begin by moving 20 coins from the 4th pile to the 3rd which gives the following configuration: 0, 20, 50, 20, 50. Regardless of what player B does, player A should follow by making sure the 2nd and 4th piles have the same number of coins. For example, if B moves 12 coins from the 3rd pile to the 2nd to get 0, 32, 38, 20, 50 player A could move 12 coins from the 5th pile to the 4th or move 12 coins from the 2nd pile to the 1st. Then B makes his next move and player A follows by ensuring the 2nd and 4th piles have the same number of coins. In particular, this ensures B's moves will always result in the 2nd and 4th piles having different numbers of coins (why?) If B could win then his finishing move would have the 2nd and 4th piles with the same number of coins (0)--but this is impossible! Thus this strategy ensures that A will win.
Note: There are still some details that I glossed over, but I'll leave these for the reader to work out.