AMC12 2013 B

AMC12 2013 B · Q18

AMC12 2013 B · Q18. It mainly tests Games (basic), Remainders & modular arithmetic.

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 和 Jenna 玩下面这个轮流进行的游戏。桌上有若干枚硬币。轮到 Barbara 时，她必须拿走 2 枚或 4 枚硬币；但如果桌上只剩 1 枚硬币，则她这一回合无法行动并失去回合。轮到 Jenna 时，她必须拿走 1 枚或 3 枚硬币。由掷硬币决定谁先手。拿走最后一枚硬币的人赢得游戏。假设双方都采用最优策略：当游戏开始时有 2013 枚硬币，以及当开始时有 2014 枚硬币，分别是谁会获胜？

(A) Barbara will win with 2013 coins, and Jenna will win with 2014 coins. 2013枚时芭芭拉胜，2014枚时詹娜胜。

(B) Jenna will win with 2013 coins, and whoever goes first will win with 2014 coins. 2013枚时詹娜胜，2014枚时先手胜。

(D) Jenna will win with 2013 coins, and Barbara will win with 2014 coins. 2013枚时詹娜胜，2014枚时芭芭拉胜。

(E) Whoever goes first will win with 2013 coins, and whoever goes second will win with 2014 coins. 2013枚时先手胜，2014枚时后手胜。

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): If the game starts with 2013 coins and Jenna starts, then she picks 3 coins, and then no matter how many Barbara chooses, Jenna responds by keeping the number of remaining coins congruent to 0 (mod 5). That is, she picks 3 if Barbara picks 2, and she picks 1 if Barbara picks 4. This ensures that on her last turn Jenna will leave 0 coins and thus she will win. Similarly, if Barbara starts, then Jenna can reply as before so that the number of remaining coins is congruent to 3 (mod 5). On her last turn Barbara will have 3 coins available. She is forced to remove 2 and thus Jenna will win by taking the last coin. If the game starts with 2014 coins and Jenna starts, then she picks 1 coin and reduces the game to the previous case of 2013 coins where she wins. If Barbara starts, she selects 4 coins. Then no matter what Jenna chooses, Barbara responds by keeping the number of remaining coins congruent to 0 (mod 5). This ensures that on her last turn Barbara will leave 0 coins and win the game. Thus whoever goes first will win the game with 2014 coins.

答案（B）：如果游戏开始时有 2013 枚硬币且 Jenna 先手，那么她先取 3 枚硬币；之后无论 Barbara 取多少，Jenna 都通过应对使剩余硬币数满足 $\equiv 0 \pmod{5}$。也就是说，如果 Barbara 取 2 枚，她就取 3 枚；如果 Barbara 取 4 枚，她就取 1 枚。这样可以保证在她的最后一回合，Jenna 会让硬币剩余 0 枚，从而获胜。类似地，如果 Barbara 先手，那么 Jenna 也可以像之前那样应对，使剩余硬币数满足 $\equiv 3 \pmod{5}$。在 Barbara 的最后一回合时，她将面对剩下 3 枚硬币可取，她被迫取走 2 枚，于是 Jenna 通过取走最后 1 枚硬币获胜。如果游戏开始时有 2014 枚硬币且 Jenna 先手，那么她先取 1 枚硬币，把游戏化归到前一种 2013 枚硬币的情形，从而获胜。如果 Barbara 先手，她先取 4 枚硬币。之后无论 Jenna 取多少，Barbara 都通过应对使剩余硬币数满足 $\equiv 0 \pmod{5}$。这保证在她的最后一回合，Barbara 会让硬币剩余 0 枚并赢得比赛。因此，在 2014 枚硬币的情形下，谁先手谁就会赢。

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