Difference between revisions of "2013 AMC 12B Problems/Problem 10"
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<math>\text{B(x,y)=x-3y+75}</math> | <math>\text{B(x,y)=x-3y+75}</math> | ||
| − | 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 | + | 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: |
<math>\text{1=-2x+y+75}</math> | <math>\text{1=-2x+y+75}</math> | ||
Revision as of 18:41, 3 February 2014
- The following problem is from both the 2013 AMC 12B #10 and 2013 AMC 10B #17, so both problems redirect to this page.
Contents
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 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:
Solving yields 
and 
. Since he gains one silver coin per visit to each booth, he has 
silver coins in total.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
