MMO2025 Round 4

Answer: Player 1 has a winning strategy for all initial values, including $a_0=2^{2025}$. Let us consider small values of $a_0$.

Case 1: $a_0\le 8$. To avoid repeating a number three times, each player must choose $a_n=a_{n-1}+1$. Then player 1 can increment three times in a row: $$a_0,a_0+1,a_0+2,a_0+3,$$ which forms an arithmetic progression of length 4, ending the game with a win for player 1.

Case 2: $a_0=9$. If player 1 plays $a_1=10$, then: - If player 2 plays $a_2=11$, player 1 responds with $a_3=12$ and wins via arithmetic progression. - If player 2 instead plays $a_2=S(10)=1$, the position reduces to the previously analyzed case $a_0=1$, where player 1 also wins.

Case 3: $a_0\ge 10$. We claim that using the digit sum operation $S(a)$ leads to losing positions. Note that for $a\ge 10$, we always have $$S(a)\le a-9.$$ Hence, each application of $S$ strictly decreases the value by at least 9, eventually reducing to a single-digit number, which as previously shown results in a loss for the player making that move. Therefore, under optimal play, both players avoid using $S$ and only apply $a_n=a_{n-1}+1$. Then, player 1 forces: $a_0,a_0+1,a_0+2,a_0+3$, forming an arithmetic progression of length 4, thereby winning.