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2013 AMC 12B Problems/Problem 18: Difference between revisions

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<math>\textbf{(E)}</math> Whoever goes first will win with <math>2013</math> coins, and whoever goes second will win with <math>2014</math> coins.
<math>\textbf{(E)}</math> Whoever goes first will win with <math>2013</math> coins, and whoever goes second will win with <math>2014</math> coins.


[[2013 AMC 12B Problems/Problem 18|Solution]]


==Solution==
==Solution==
== See also ==
{{AMC12 box|year=2013|ab=B|num-b=17|num-a=19}}

Revision as of 17:27, 22 February 2013

Problem

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.


Solution

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