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

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==Problem==
 
==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. When 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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A pyramid has a square base with side of length 1 and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?
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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{(A)}\ 5\sqrt{2} - 7 \qquad \textbf{(B)}\ 7 - 4\sqrt{3} \qquad \textbf{(C)}\ \frac{2\sqrt{2}}{27} \qquad \textbf{(D)}\ \frac{\sqrt{2}}{9} \qquad \textbf{(E)}\ \frac{\sqrt{3}}{9}</math>
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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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==Solution==
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We spit into 2 cases: 2013 coins, and 2014 coins.
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<math>\textbf{2013 coins:}</math>
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Notice that when there are <math>5</math> 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 <math>3</math> coins. Then, whenever Barbara takes coins, Jenna will take the amount that makes the total coins taken in that round <math>5</math>. (For instance, if Barbara takes <math>4</math> coins, Jenna will take <math>1</math>). Eventually, since <math>2010=0 (\text{mod }5)</math> it will be Barbara's move with <math>5</math> 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 <math>5</math>. Since <math>2013=3 (\text{mod }5)</math>, it will be Barbara's move with <math>3</math> coins remaining, so she will have to take <math>2</math> coins, allowing Jenna to take the last coin. Therefore, Jenna will win with <math>2013</math> coins.
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<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 }5)</math>, it will be Jenna's turn with <math>5</math> coins left, so
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Barbara will win. In this case, whoever moves first wins.
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Based on this, the answer is <math>\boxed{\textbf{(B)}}</math>
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== See also ==
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{{AMC12 box|year=2013|ab=B|num-b=17|num-a=19}}
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{{MAA Notice}}

Revision as of 18:34, 28 August 2020

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. When 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 to 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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