AMC12 2021 B

We say that a game state is an N-position if it is winning for the next player (the player to move), and a P-position if it is winning for the other player. We are trying to find which of the given states is a P-position. First we note that symmetrical positions are P-positions, as the second player can win by mirroring the first player's moves. It follows that $(6, 1, 1)$ is an N-position, since we can win by moving to $(2, 2, 1, 1)$; this rules out $\textbf{(A)}$. We next look at $(6, 2, 1)$. The possible next states are \[(6, 2), (6, 1, 1), (6, 1), (5, 2, 1), (4, 2, 1, 1),(4, 2, 1), (3, 2, 2, 1), (3, 2, 1, 1), (2, 2, 2, 1).\] None of these are symmetrical, so we might reasonably suspect that they are all N-positions. Indeed, it just so happens that for all of these states except $(6, 2)$ and $(6, 1)$, we can win by moving to $(2, 2, 1, 1)$; it remains to check that $(6, 2)$ and $(6, 1)$ are N-positions. To save ourselves work, it would be nice if we could find a single P-position directly reachable from both $(6, 2)$ and $(6, 1)$. We notice that $(3, 2, 1)$ is directly reachable from both states, so it would suffice to show that $(3, 2, 1)$ is a P-position. Indeed, the possible next states are \[(3, 2), (3, 1, 1), (3, 1), (2, 2, 1), (2, 1, 1, 1), (2, 1, 1),\] which allow for the following refutations: \begin{align*} &(3, 2) \to (2, 2), && &&(3, 1, 1) \to (1, 1, 1, 1), && &&(3, 1) \to (1, 1), \\ &(2, 2, 1) \to (2, 2), && &&(2, 1, 1, 1) \to (1, 1, 1, 1), && &&(2, 1, 1) \to (1, 1). \end{align*} Hence, $(3, 2, 1)$ is a P-position, so $(6, 2)$ and $(6, 1)$ are both N-positions, along with all other possible next states from $(6, 2, 1)$ as noted before. Thus, $(6, 2, 1)$ is a P-position, so our answer is $\boxed{\textbf{(B)}}$. (For completeness, we could also rule out $\textbf{(C)}$, $\textbf{(D)}$ and $\textbf{(E)}$ as in Solution 2.)