# Problem #2290

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 2290 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. What 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? $\textbf{(A)}$ 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. $\textbf{(E)}$ Whoever goes first will win with $2013$ coins, and whoever goes second will win with $2014$ coins. This problem is copyrighted by the American Mathematics Competitions.
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• Reduce fractions to lowest terms and enter in the form 7/9.
• Numbers involving pi should be written as 7pi or 7pi/3 as appropriate.
• Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate.
• Exponents should be entered in the form 10^10.
• If the problem is multiple choice, enter the appropriate (capital) letter.
• Enter points with parentheses, like so: (4,5)
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