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Difference between revisions of "2013 AMC 12B Problems/Problem 18"
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<math>\textbf{2014 coins:}</math>
 
<math>\textbf{2014 coins:}</math>
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If Jenna moves first, she will take <math>1</math> coin, leaving <math>2013</math> coins, and she wins as shown above. If Barbara moves first, she can take <math>4</math> coins, leaving <math>2010</math>. After every move by Jenna, Barbara will then take the number of coins that makes the total taken in that round <math>5</math>. Since <math>2010=0\text{(mod }3)</math>, it will be Jenna's turn with <math>5</math> coins left, so  
+
If Jenna moves first, she will take <math>1</math> coin, leaving <math>2013</math> coins, and she wins as shown above. If Barbara moves first, she can take <math>4</math> coins, leaving <math>2010</math>. After every move by Jenna, Barbara will then take the number of coins that makes the total taken in that round <math>5</math>. Since <math>2010=0\text{(mod }5)</math>, it will be Jenna's turn with <math>5</math> coins left, so  
 
Barbara will win. In this case, whoever moves first wins.
 
Barbara will win. In this case, whoever moves first wins.
  

Revision as of 17:57, 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

We spit into 2 cases: 2013 coins, and 2014 coins.

$\textbf{2013 coins:}$ Notice that when there are $5$ coins left, whoever moves first loses, as they must leave an amount of coins the other person can take. If Jenna goes first, she can take $3$ coins. Then, whenever Barbara takes coins, Jenna will take the amount that makes the total coins taken in that round $5$. (For instance, if Barbara takes $4$ coins, Jenna will take $1$). Eventually, since $2010=0 (\text{mod }5)$ it will be Barbara's move with $5$ coins remaining, so she will lose. If Barbara goes first, on each round, Jenna will take the amount of coins that makes the total coins taken on that round $5$. Since $2013=3 (\text{mod }5)$, it will be Barbara's move with $3$ coins remaining, so she will have two take $2$ coins, allowing Jenna to take the last coin. Therefore, Jenna will win with $2013$ coins.

$\textbf{2014 coins:}$ If Jenna moves first, she will take $1$ coin, leaving $2013$ coins, and she wins as shown above. If Barbara moves first, she can take $4$ coins, leaving $2010$. After every move by Jenna, Barbara will then take the number of coins that makes the total taken in that round $5$. Since $2010=0\text{(mod }5)$, it will be Jenna's turn with $5$ coins left, so Barbara will win. In this case, whoever moves first wins.

Based on this, the answer is $\boxed{\textbf{(B)}}$

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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