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2013 AMC 12B Question 18

AMC12 · 2013 AMC 12B

Question 18

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?
  1. Barbara will win with $2013$ coins, and Jenna will win with
    $2014$ coins.
  2. Jenna will win with $2013$ coins, and whoever
    goes first will win with $2014$ coins.
  3. Barbara will win with $2013$ coins, and whoever goes second
    will win with $2014$ coins.
  4. Jenna will win with $2013$ coins, and Barbara will win with
    $2014$ coins.
  5. Whoever goes first will win with $2013$ coins, and whoever
    goes second will win with $2014$ coins.