OMCC 2009 Problem 3

There are 2009 boxes numbered from 1 to 2009, some of which contain stones. Two players, $A$ and $B$ , play alternately, starting with $A$ . A move consists in selecting a non-empty box $i$ , taking one or more stones from that box and putting them in box $i + 1$ . If $i = 2009$ , the selected stones are eliminated. The player who removes the last stone wins a) If there are 2009 stones in the box 2 and the others are empty, find a winning strategy for either player. b) If there is exactly one stone in each box, find a winning strategy for either player.