Difference between revisions of "2013 AMC 12B Problems/Problem 18"
| Line 12: | Line 12: | ||
<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. | ||
| − | |||
==Solution== | ==Solution== | ||
| − | |||
| − | |||
| − | |||
Revision as of 18:27, 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 
or 
coins, unless only one coin remains, in which case she loses her turn. What it is Jenna’s turn, she must remove 
or 
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 
coins and when the game starts with 
coins?
Barbara will win with coins and Jenna will win with coins.
Jenna will win with coins, and whoever goes first will win with coins.
Barbara will win with coins, and whoever goes second will win with coins.
Jenna will win with coins, and Barbara will win with coins.
Whoever goes first will win with coins, and whoever goes second will win with coins.
