Question

Barbara 和 Jenna 轮流进行如下游戏。桌上放着若干枚硬币。轮到 Barbara 时，她必须拿走 $2$ 枚或 $4$ 枚硬币；如果只剩一枚硬币，她就跳过这一轮。轮到 Jenna 时，她必须拿走 $1$ 枚或 $3$ 枚硬币。由抛硬币决定谁先走，拿走最后一枚硬币的人获胜。假设两人都采用最佳策略。当游戏分别从 $2013$ 枚和 $2014$ 枚硬币开始时，谁会获胜？

Barbara and Jenna play the following game, in which they take turns. A number of coins lie on a table. When it is Barbara’s turn, she must remove $2$ or $4$ coins, unless only one coin remains, in which case she loses her turn. When it is Jenna’s turn, she must remove $1$ or $3$ coins. A coin flip determines who goes first. Whoever removes the last coin wins the game. Assume both players use their best strategy. Who will win when the game starts with $2013$ coins and when the game starts with $2014$ coins?

Barbara 会在 $2013$ 枚硬币时获胜，Jenna 会在 $2014$ 枚硬币时获胜。

Barbara will win with $2013$ coins, and Jenna will win with $2014$ coins.

Jenna 会在 $2013$ 枚硬币时获胜，而 $2014$ 枚硬币时先手获胜。

Jenna will win with $2013$ coins, and whoever goes first will win with $2014$ coins.

Barbara 会在 $2013$ 枚硬币时获胜，而 $2014$ 枚硬币时后手获胜。

Barbara will win with $2013$ coins, and whoever goes second will win with $2014$ coins.

Jenna 会在 $2013$ 枚硬币时获胜，Barbara 会在 $2014$ 枚硬币时获胜。

Jenna will win with $2013$ coins, and Barbara will win with $2014$ coins.

$2013$ 枚硬币时先手获胜， $2014$ 枚硬币时后手获胜。

Whoever goes first will win with $2013$ coins, and whoever goes second will win with $2014$ coins.

答案：B

知识点：组合游戏不变量模运算

难度评级：2070

小提示：

追踪硬币数量模 $5$ 的余数。

Track the number of coins modulo $5$

大提示：

Jenna 可以总是在自己走完后恢复到 $5$ 的倍数：Barbara 拿 $2$ 时她拿 $3$ ，Barbara 拿 $4$ 时她拿 $1$ ；再判断每个初始数量下谁能维持不变量。

Jenna can always restore a multiple of $5$ after her move, answering Barbara’s $2$ with $3$ and her $4$ with $1;$ decide who can maintain the invariant for each starting count

解答：

按模 $5$ 分析。因为 $2013 \equiv 3$ ，无论谁先走，Jenna 都能获胜。如果 Jenna 先走，她先拿走 $3$ 枚，使剩余数量成为 $5$ 的倍数；此后 Barbara 拿走 $2$ 枚时，她就拿走 $3$ 枚，Barbara 拿走 $4$ 枚时，她就拿走 $1$ 枚，从而始终留下 $5$ 的倍数，并最终拿走最后一枚。如果 Jenna 后走，她可以使每轮结束后的硬币数保持 $\equiv 3 \pmod 5$ ，直到 Barbara 面对 $3$ 枚硬币，只能拿走 $2$ 枚，把最后一枚留给 Jenna。因为 $2014 \equiv 4$ ，这时先手获胜：Jenna 先走可把局面化为 $2013$ 枚的情形；Barbara 先走则可先拿走 $4$ 枚，此后维持 $5$ 的倍数。因此选择 B。所以正确答案是 B。

Work modulo $5.$ With $2013 \equiv 3$ coins, Jenna wins either way: going first she takes $3$ to leave a multiple of $5,$ then answers Barbara’s $2$ with $3$ and $4$ with $1$ to keep multiples of $5,$ eventually taking the last coin; going second she keeps the count $\equiv 3 \pmod 5$ until Barbara is stuck at $3$ coins, must remove $2,$ and leaves Jenna the last coin. With $2014 \equiv 4$ coins, whoever goes first wins: Jenna first reduces to the $2013$ case, while Barbara first takes $4$ and then keeps multiples of $5.$ This is choice B. Thus, the correct answer is B.

2013 AMC 12B 第 18 题

答案：B

难度评级：2070

小提示：

大提示：

解答：

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