Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 2)
 
(5 intermediate revisions by 4 users not shown)
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!==
 
  
−
We spit into 2 cases: 2013 coins, and 2014 coins.
+
 
 +
==Solution 1==
 +
 
 +
We split into 2 cases: 2013 coins, and 2014 coins.
  
 
<math>\textbf{2013 coins:}</math>
 
<math>\textbf{2013 coins:}</math>
Line 24: Line 26:
  
 
Based on this, the answer is <math>\boxed{\textbf{(B)}}</math>
 
Based on this, the answer is <math>\boxed{\textbf{(B)}}</math>
 +
 +
 +
==Solution 2==
 +
 +
We start with some key observations.
 +
 +
First, notice that when there are <math>0 (\text{mod }5)</math> coins left, whoever moves first loses, since the other player can keep taking exactly <math>5</math> minus the amount the first player took, eventually removing the last coin.
 +
 +
Furthermore, notice that if Barbara is going first when there are <math>1 (\text{mod }5)</math> coins left, she will lose, since the Jenna can follow a similar strategy to the <math>0 (\text{mod }5)</math> case. This will leave Barbara with only 1 coin at the end, in which she loses her turn and Jenna wins.
 +
 +
 +
 +
With these in mind, we split into 2 cases: 2013 coins, and 2014 coins.
 +
 +
 +
<math>\textbf{2013 coins:}</math>
 +
 +
If Barbara goes first, she can remove either <math>2</math> or <math>4</math> coins. If she removes <math>2</math> coins, then Jenna will remove <math>1</math> coin, so there will be <math>0 (\text{mod }5)</math> left (hence Barbara will lose). Alternatively, if she removes <math>4</math> coins, then Jenna will remove <math>3</math> coins, so there will be <math>1 (\text{mod }5)</math> left (and Barbara will lose again). In both of these cases, Jenna will win.
 +
 +
 +
If Jenna goes first, she will remove <math>3</math> coins, so there will be <math>0 (\text{mod }5)</math> coins left. Hence Jenna will win if she goes first.
 +
 +
 +
Putting these together, Jenna will always win with 2013 coins.
 +
 +
 +
 +
<math>\textbf{2014 coins:}</math>
 +
 +
If Barbara goes first, she will remove <math>4</math> coins, so there will be <math>0 (\text{mod }5)</math> coins left. Hence Barbara will win.
 +
 +
 +
Or, if Jenna goes first, she will remove <math>3</math> coins, so there will be <math>1 (\text{mod }5)</math> coins left. Hence Jenna will win.
 +
 +
 +
Based on these, whoever goes first will win with 2014 coins.
 +
 +
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(B)}}</math>.
 +
 +
~xHypotenuse
 +
 +
==Video Solution==
 +
https://youtu.be/I6JgSPbL6gY  ~Math Problem Solving Skills
 +
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2013|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:59, 16 July 2024

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 1

We split 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 $\boxed{\textbf{(B)}}$


Solution 2

We start with some key observations.

First, notice that when there are $0 (\text{mod }5)$ coins left, whoever moves first loses, since the other player can keep taking exactly $5$ minus the amount the first player took, eventually removing the last coin.

Furthermore, notice that if Barbara is going first when there are $1 (\text{mod }5)$ coins left, she will lose, since the Jenna can follow a similar strategy to the $0 (\text{mod }5)$ case. This will leave Barbara with only 1 coin at the end, in which she loses her turn and Jenna wins.


With these in mind, we split into 2 cases: 2013 coins, and 2014 coins.


$\textbf{2013 coins:}$

If Barbara goes first, she can remove either $2$ or $4$ coins. If she removes $2$ coins, then Jenna will remove $1$ coin, so there will be $0 (\text{mod }5)$ left (hence Barbara will lose). Alternatively, if she removes $4$ coins, then Jenna will remove $3$ coins, so there will be $1 (\text{mod }5)$ left (and Barbara will lose again). In both of these cases, Jenna will win.


If Jenna goes first, she will remove $3$ coins, so there will be $0 (\text{mod }5)$ coins left. Hence Jenna will win if she goes first.


Putting these together, Jenna will always win with 2013 coins.


$\textbf{2014 coins:}$

If Barbara goes first, she will remove $4$ coins, so there will be $0 (\text{mod }5)$ coins left. Hence Barbara will win.


Or, if Jenna goes first, she will remove $3$ coins, so there will be $1 (\text{mod }5)$ coins left. Hence Jenna will win.


Based on these, whoever goes first will win with 2014 coins.


Therefore, the answer is $\boxed{\textbf{(B)}}$.

~xHypotenuse

Video Solution

https://youtu.be/I6JgSPbL6gY ~Math Problem Solving Skills


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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_18&oldid=222753"