Question

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

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

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

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

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

Solution:

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 Problem 18

Solution:

Problem 18 in Other Years