2013 AMC 12B Problems/Problem 10
- The following problem is from both the 2013 AMC 12B #10 and 2013 AMC 10B #17, so both problems redirect to this page.
Contents
[hide]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 1
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
Solution 2
Let denote the number of visits to the first booth and denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows: There are no legal exchanges when he has fewer than red coins and fewer than blue coins, namely when he has red coin and blue coins. We can then create a system of equations: Solving yields and . Since he gains one silver coin per visit to each booth, he has 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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