Difference between revisions of "2013 AMC 12B Problems/Problem 18"
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==Problem== | ==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 <math>2</math> or <math>4</math> coins, unless only one coin remains, in which case she loses her turn. | + | 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? |
<math> \textbf{(A)}</math> Barbara will win with <math>2013</math> coins and Jenna will win with <math>2014</math> coins. | <math> \textbf{(A)}</math> Barbara will win with <math>2013</math> coins and Jenna will win with <math>2014</math> coins. | ||
Line 13: | Line 13: | ||
− | ==Solution== | + | |
+ | ==Solution 1== | ||
+ | |||
+ | We split into 2 cases: 2013 coins, and 2014 coins. | ||
+ | |||
+ | <math>\textbf{2013 coins:}</math> | ||
+ | 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. | ||
+ | |||
+ | <math>\textbf{2014 coins:}</math> | ||
+ | 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. | ||
+ | |||
+ | 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}} |
Latest revision as of 11:59, 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 or
coins, unless only one coin remains, in which case she loses her turn. When 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.
Solution 1
We split into 2 cases: 2013 coins, and 2014 coins.
Notice that when there are
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
coins. Then, whenever Barbara takes coins, Jenna will take the amount that makes the total coins taken in that round
. (For instance, if Barbara takes
coins, Jenna will take
). Eventually, since
it will be Barbara's move with
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
. Since
, it will be Barbara's move with
coins remaining, so she will have to take
coins, allowing Jenna to take the last coin. Therefore, Jenna will win with
coins.
If Jenna moves first, she will take
coin, leaving
coins, and she wins as shown above. If Barbara moves first, she can take
coins, leaving
. After every move by Jenna, Barbara will then take the number of coins that makes the total taken in that round
. Since
, it will be Jenna's turn with
coins left, so
Barbara will win. In this case, whoever moves first wins.
Based on this, the answer is
Solution 2
We start with some key observations.
First, notice that when there are coins left, whoever moves first loses, since the other player can keep taking exactly
minus the amount the first player took, eventually removing the last coin.
Furthermore, notice that if Barbara is going first when there are coins left, she will lose, since the Jenna can follow a similar strategy to the
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.
If Barbara goes first, she can remove either or
coins. If she removes
coins, then Jenna will remove
coin, so there will be
left (hence Barbara will lose). Alternatively, if she removes
coins, then Jenna will remove
coins, so there will be
left (and Barbara will lose again). In both of these cases, Jenna will win.
If Jenna goes first, she will remove coins, so there will be
coins left. Hence Jenna will win if she goes first.
Putting these together, Jenna will always win with 2013 coins.
If Barbara goes first, she will remove coins, so there will be
coins left. Hence Barbara will win.
Or, if Jenna goes first, she will remove coins, so there will be
coins left. Hence Jenna will win.
Based on these, whoever goes first will win with 2014 coins.
Therefore, the answer is .
~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.