Difference between revisions of "2013 AMC 12B Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | <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> |
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=B|num-b=9|num-a=11}} | {{AMC12 box|year=2013|ab=B|num-b=9|num-a=11}} | ||
Revision as of 10:46, 7 April 2013
- 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 
red tokens and 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?
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 
See also
| 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 | |
