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Difference between revisions of "2013 AMC 12B Problems/Problem 10"
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If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first combination Combo 1, and the second combination Combo 2.
 
If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first combination Combo 1, and the second combination Combo 2.
  
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In other words, Alex can exchange 5 red tokens for 4 silver tokens as long as he has at least 6 red tokens, and Alex can exchange 5 blue tokens for 3 silver tokens as long as he has at least 6 blue tokens. Hence doing as many Combo 1 and Combo 2's as possible, we end up with 5 red, 5 blue, and 98 silver tokens.
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In other words, Alex can exchange 5 red tokens for 4 silver tokens as long as he has at least 6 red tokens, and Alex can exchange 5 blue tokens for 3 silver tokens as long as he has at least 6 blue tokens. Hence after performing 14 Combo 1's and 14 Combo 2's, we end up with 5 red, 5 blue, and 98 silver tokens.
  
 
Finally, Alex can visit the blue booth once, then do Combo 1, then visit the blue booth once more to end up with 1 red token, 2 blue tokens, and <math>\boxed{\textbf{(E)}\ 103}</math> silver tokens, at which point it is clear he cannot use the booths anymore.
 
Finally, Alex can visit the blue booth once, then do Combo 1, then visit the blue booth once more to end up with 1 red token, 2 blue tokens, and <math>\boxed{\textbf{(E)}\ 103}</math> silver tokens, at which point it is clear he cannot use the booths anymore.
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We can approach this problem by assuming he goes to the red booth first. You start with <math>75 \text{R}</math> and <math>75 \text{B}</math> and at the end of the first booth, you will have <math>1 \text{R}</math> and <math>112 \text{B}</math> and <math>37 \text{S}</math>. We now move to the blue booth, and working through each booth until we have none left, we will end up with:<math>1 \text{R}</math>, <math>2 \text{B}</math> and <math>103 \text{S}</math>. So, the answer is <math>\boxed{\textbf{(E)}103}</math>
 
We can approach this problem by assuming he goes to the red booth first. You start with <math>75 \text{R}</math> and <math>75 \text{B}</math> and at the end of the first booth, you will have <math>1 \text{R}</math> and <math>112 \text{B}</math> and <math>37 \text{S}</math>. We now move to the blue booth, and working through each booth until we have none left, we will end up with:<math>1 \text{R}</math>, <math>2 \text{B}</math> and <math>103 \text{S}</math>. So, the answer is <math>\boxed{\textbf{(E)}103}</math>
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==Solution 3==
 
==Solution 3==

Revision as of 00:08, 2 January 2019

The following problem is from both the 2013 AMC 12B #10 and 2013 AMC 10B #17, so both problems redirect to this page.

Problem

Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end?

$\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103$

Solution 1

If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first combination Combo 1, and the second combination Combo 2.

In other words, Alex can exchange 5 red tokens for 4 silver tokens as long as he has at least 6 red tokens, and Alex can exchange 5 blue tokens for 3 silver tokens as long as he has at least 6 blue tokens. Hence after performing 14 Combo 1's and 14 Combo 2's, we end up with 5 red, 5 blue, and 98 silver tokens.

Finally, Alex can visit the blue booth once, then do Combo 1, then visit the blue booth once more to end up with 1 red token, 2 blue tokens, and $\boxed{\textbf{(E)}\ 103}$ silver tokens, at which point it is clear he cannot use the booths anymore.

Solution 2==

We can approach this problem by assuming he goes to the red booth first. You start with $75 \text{R}$ and $75 \text{B}$ and at the end of the first booth, you will have $1 \text{R}$ and $112 \text{B}$ and $37 \text{S}$. We now move to the blue booth, and working through each booth until we have none left, we will end up with:$1 \text{R}$, $2 \text{B}$ and $103 \text{S}$. So, the answer is $\boxed{\textbf{(E)}103}$

Solution 3

Let $x$ denote the number of visits to the first booth and $y$ denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows: \[R(x,y)=-2x+y+75\] \[B(x,y)=x-3y+75\] There are no legal exchanges when he has fewer than $2$ red coins and fewer than $3$ blue coins, namely when he has $1$ red coin and $2$ blue coins. We can then create a system of equations: \[1=-2x+y+75\] \[2=x-3y+75\] Solving yields $x=59$ and $y=44$. Since he gains one silver coin per visit to each booth, he has $x+y=44+59=\boxed{\textbf{(E)}103}$ silver coins in total.

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 10 Problems and Solutions
2013 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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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