Difference between revisions of "2013 AMC 10B Problems/Problem 17"
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==Solution== | ==Solution== | ||
− | 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> | + | 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> |
Revision as of 08:18, 26 February 2013
Problem
Alex has red tokens and 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?
Solution
We can approach this problem by assuming he goes to the red booth first. You start with and and at the end of the first booth, you will have and and . We now move to the blue booth, and working through each booth until we have none left, we will end up with:, and . So, the answer is