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Difference between revisions of "2013 AMC 10B Problems/Problem 17"

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==Solution==
 
==Solution==
  
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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>.
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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>\textbf{(E)}</math>

Revision as of 08:17, 26 February 2013

Problem

Alex has $75$ red tokens and $75$ blue tokens. There is a booth where Alex can give two red tokens and recieve in return a silver token and a blue token, and another booth where Alex can give three blue tokens and recieve 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

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 $\textbf{(E)}$

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