Divisor Game - Problem Statement

Start with $n$. Each turn, pick divisor $x$ of current number where $0 < x < n$, subtract it. Can't move = lose. $dp[1] = false$ (no valid divisor). $dp[2] = true$ (pick $1$, opponent gets 1, loses).

Transition: $dp[n] = \bigvee_{x | n, x < n} \lnot dp[n-x]$. Check all proper divisors. Pattern: $dp[n] = (n \mod 2 == 0)$. Even numbers win, odd numbers lose. Proof: even can always subtract 1 (a divisor), odd - 1 = even, opponent wins.