[Concyclicboy]

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

[gilcu3]

An odd box is, for example, the box 17, the box 35,...

For the part, the player win because he allways can play such that all the stones, after he play, will be in even boxes, and then the player cannot win, and so like the game is finite, the player must win.

For the part, I use the same strategy, it is that the player who can play such that, after the other player plays, the number of boxes odd and nonempty will be odd, and then is at least one, so he cannot lost. I will prove that the player win.
Let be the number of odd boxes wich ar nonempty in any moment.
In the begining of the game there are 1005 boxes that are nonempty and odd. The first play of the player is select one stone of the box 2009. Now is even. Then the player B can play of two ways:
1- If he select stones from the box , then in the next play the player select stones from the box (it is easy to see that it allways is possible).
2- If he select stone from the box , then in the next play the player select stone from the box .
We see that like in the begining there is one stone in each box, after plays, the number is allways odd if played 2 (so allways exist).
After plays the number is even and in each odd box the number of stones is 0 or 1. So the player allways can play, and then he must win.

[Dumel]

gilcu3 wrote:

Let be the number of odd boxes wich ar nonempty in any moment.
In the begining of the game there are 1005 boxes that are nonempty and odd. The first play of the player is select one stone of the box 2009. Now is even. Then the player B can play of two ways:
1- If he select stones from the box , then in the next play the player select stones from the box (it is easy to see that it allways is possible).

if the first play of B is 1., then k is even after the play of A

[gilcu3]

Yes, I proved that after each play of , the number is even.