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There are two players. We mark them I and II. Player I says any positive divisor of the number $10^{2n}$. Thereafter, the players alternate turns with player II starting. In each turn, a player can multiply or divide by a prime number ($2$ or $5$) the integer number that another player said during the previous turn so that the result is a positive divisor of the number $10^{2n}$ that players did not say before. A player who cannot move is considered to lose the game. Which player has a winning strategy?

My work. Most likely the player I has a winning strategy. I proved that this problem is equivalent to the following problem: "Two players move "rook" on the chessboard $(2n+1) \times (2n+1)$. In one move, the "rook" can be moved horizontally or vertically to just one square. The player I chooses the starting position of the "rook" on the chessboard. Thereafter, the players alternate moves with player II starting. In each move, the player moves the "rook" to the square, in which she had never been before. A player who cannot move is considered to lose the game. Which player has a winning strategy?"

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    $\begingroup$ What you have said looks sensible: player 1 can choose where to start and the maximum number of positions is odd, which combined should give player 1 an overwhelming advantage. With $n=1$, playing in the centre (i.e. saying $10$) is a guaranteed win for player 1 $\endgroup$ – Henry Jun 26 at 17:56
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Your reduction is nice. Base on it, one can prove that the first player can win as follows:

Color the chessboard into black and white such that adjacent cells have different colors. Then one color occurred once more than the other, suppose white occurred more.

The first player starts at any white square (equivalently, pick the initial number of the form $a^2 \cdot 10^k$). Pair the remaining squares into $1\times2$ dominoes. Wherever the second player moves, it must lie on some domino. The first player then immediately moves to the adjacent tile in the domino. This guarantees that the first player always have a move.

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  • $\begingroup$ Very nice solution! Thank you very much! $\endgroup$ – Witold Jun 26 at 21:28

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