AMC 12 2013b Problem 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?

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

\textbf{(A)}$Barbara will win with$2013$coins and Jenna will win with$2014$coins.$\textbf{(B)}$Jenna will win with$2013$coins, and whoever goes first will win with$2014$coins.$\textbf{(C)}$Barbara will win with$2013$coins, and whoever goes second will win with$2014$coins.$\textbf{(D)}$Jenna will win with$2013$coins, and Barbara will win with$2014$coins.$\textbf{(E)}