Rockies

Suppose that Alice is faced with a pile of $N$ stones, where $N \equiv 0,2 \pmod{7}$ . The following table shows that, no matter what move Alice makes, Bob can make a responding move which ensures that Alice is faced with a pile of $M$ stones, where $M \equiv 0,2 \pmod{7}$ again. $\begin{array}{cccc} N \equiv ? \pmod{7} & \mathrm{Alice} & \mathrm{Bob} & M \equiv ? \pmod{7} \\ \hline 0 & 1 & 4 & 2 \\ 0 & 3 & 4 & 0 \\ 0 & 4 & 3 & 0 \\ 2 & 1 & 1 & 0 \\ 2 & 3 & 4 &2 \\ 2 & 4 & 3 & 2 \end{array}$ In addition, if Alice is faced with a pile of $N$ stones, where $N \equiv 0,2 \pmod{7}$ , then it is not possible for Alice to leave a multiple of $7$ stones (so that it is impossible for Alice to take the last stone and win). Since any player can always remove a single stone, a stalemate is not possible. Since $2018 \equiv 2 \pmod{7}$ , we deduce that Bob must always win.

Alternatively, we can show that the Sprague-Grundy function for this game is $G$ , where $G(n) \; = \; \left\{ \begin{array}{lll} 0 & \hspace{1cm} & n \equiv 0,2 \pmod{7} \\ 1 & & n \equiv 1,3 \pmod{7} \\ 2 && n \equiv 4,6 \pmod{7} \\ 3 && n \equiv 5 \pmod{7} \end{array} \right.$ and so the losing positions for Alice at those for which $G(n) = 0$ , namely those for which $n \equiv 0,2 \pmod{7}$ .