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Difference between revisions of "2013 AMC 12B Problems/Problem 18"

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==Problem==
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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 <math>2</math> or <math>4</math> coins, unless only one coin remains, in which case she loses her turn. What it is Jenna’s turn, she must remove <math>1</math> or <math>3</math> 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 <math>2013</math> coins and when the game starts with <math>2014</math> coins?
  
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<math> \textbf{(A)}</math> Barbara will win with <math>2013</math> coins and Jenna will win with <math>2014</math> coins.
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<math>\textbf{(B)}</math> Jenna will win with <math>2013</math> coins, and whoever goes first will win with <math>2014</math> coins.
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<math>\textbf{(C)}</math> Barbara will win with <math>2013</math> coins, and whoever goes second will win with <math>2014</math> coins.
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<math>\textbf{(D)}</math> Jenna will win with <math>2013</math> coins, and Barbara will win with <math>2014</math> coins.
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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.
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[[2013 AMC 12B Problems/Problem 18|Solution]]
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==Solution==
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== See also ==
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{{AMC12 box|year=2013|ab=B|num-b=17|num-a=19}}

Revision as of 18:12, 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

Solution

See also

2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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