Question

Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes $1$ token or $4$ tokens from the stack. The player who removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ such that there is a strategy that guarantees that Bob wins, regardless of Alice's moves.

Answer

Answer: 809

Concepts:combinatorial game modular arithmetic small cases

Difficulty rating: 2110

Solution:

Call $n$ a losing position if the player about to move loses with best play. We claim the losing positions are exactly $n \equiv 0$ or $2 \pmod 5.$ From such an $n,$ removing $1$ or $4$ tokens leaves $n \equiv 4, 1, 3 \pmod 5$ — never again $0$ or $2$ — while from any $n \equiv 1, 3, 4 \pmod 5$ one move reaches a position $\equiv 0$ or $2 \pmod 5$ (remove $1,$ $1,$ $4$ tokens respectively). Since $n = 0$ is a loss for the player to move, induction confirms the pattern.

Bob wins exactly when $n$ is a losing position for Alice. Among $1 \le n \le 2024$ there are $404$ multiples of $5$ and $405$ values $n \equiv 2 \pmod 5$ (from $2$ to $2022$ ), for a total of $404 + 405 = 809.$

2024 AIME I Problem 3

Answer: 809

Difficulty rating: 2110

Solution:

Problem 3 in Other Years