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Two players take alternate turns in the following game. At the outset there are two piles, containing 10,000 and 20,000 tokens, respectively. A move consists of removing any positive number of tokens from a single pile or removing $x>0$ tokens from one pile and $y>0$ tokens from the other, where $x+y$ is divisible by 2015. The player who cannot make a move loses. Which player has a winning strategy?
The first player wins.

He should present his opponent with one of the following positions:

$$
(0,0), \quad(1,1), \quad(2,2), \quad \ldots, \quad(2014,2014) .
$$

All these positions have different total numbers of tokens modulo 2015. Therefore, if the game starts from two piles of arbitrary sizes, it is possible to obtain one of these positions just by the first move. In our case

$$
10,000+20,000 \equiv 1790 \bmod 2015,
$$

and the first player can leave to his opponent the position $(895,895)$.

Now the second type of move can no longer be carried out. If the second player removes $n$ tokens from one pile, the first player may always respond be removing $n$ tokens from the other pile.

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