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2013 AMC 12B Problem 10

Problem 10 of 25EasierAlgebraProblem-Solving Techniques

Alex has 7575 red tokens and 7575 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?

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Solution

After mm red-booth and nn blue-booth exchanges, Alex has 75−(2m−n)75 - (2m - n) red tokens, 75−(3n−m)75-(3n-m) blue tokens, and m+nm+n silver tokens. At termination the red count is 00 or 1,1, and the blue count is 0,1,0,1, or 2.2. Solving the token equations over these six cases leaves only (m,n)=(59,44),(m,n)=(59,44), ending at (1,2),(1,2), or (m,n)=(60,45),(m,n)=(60,45), ending at (0,0).(0,0). The latter is unreachable: its final exchange would have to start at either (−1,3)(-1,3) or (2,−1).(2,-1). Hence Alex finishes with 59+44=10359+44=103 silver tokens. Thus, the correct answer is E.
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Tagged: system of equations · invariant

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