Difference between revisions of "2013 AMC 12B Problems/Problem 10"

Revision as of 15:45, 1 October 2020

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 $75$ red tokens and $75$ 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?

$\textbf{(A)}\ 62 \qquad \textbf{(B)}\ 82 \qquad \textbf{(C)}\ 83 \qquad \textbf{(D)}\ 102 \qquad \textbf{(E)}\ 103$

Solution 1

If Alex goes to the red booth 3 times, then goes to the blue booth once, Alex can exchange 6 red tokens for 4 silver tokens and one red token. Similarly, if Alex goes to the blue booth 2 times, then goes to the red booth once, Alex can exchange 6 blue tokens for 3 silver tokens and one blue token. Let's call the first combination Combo 1, and the second combination Combo 2.

In other words, Alex can exchange 5 red tokens for 4 silver tokens as long as he has at least 6 red tokens, and Alex can exchange 5 blue tokens for 3 silver tokens as long as he has at least 6 blue tokens. Hence after performing 14 Combo 1's and 14 Combo 2's, we end up with 5 red, 5 blue, and 98 silver tokens.

Finally, Alex can visit the blue booth once, then do Combo 1, then visit the blue booth once more to end up with 1 red token, 2 blue tokens, and $\boxed{\textbf{(E)}\ 103}$ silver tokens, at which point it is clear he cannot use the booths anymore.

Solution 2

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 $\boxed{\textbf{(E)}103}$

Solution 3

Let $x$ denote the number of visits to the first booth and $y$ denote the number of visits to the second booth. Then we can describe the quantities of his red and blue coins as follows: $R(x,y)=-2x+y+75$ $B(x,y)=x-3y+75$ 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: $1=-2x+y+75$ $2=x-3y+75$ Solving yields $x=59$ and $y=44$ . Since he gains one silver coin per visit to each booth, he has $x+y=44+59=\boxed{\textbf{(E)}103}$ silver coins in total.

Solution 4 (Invariance)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 ==Solution 4 (Invariance)==
-Suppose we set the values of the coins: Red = 4, Blue = 3, Silver = 5. This way, whenever an exchange is made, Alex's total value in coins does not change. His total value at the beginning (and thus also at the end) is <math>3\cdot 75+4\cdot 75=525</math>, so his theoretical possible total (if everything works out perfectly) is 105.
-Note that the maximum possible leftover red coins is 1, and maximum possible leftover blue coins is 2, so the maximum leftover non-silver value is 10. This perfectly gives us 103 silver coins.
 == See also ==

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