2013 AMC 12B Problem 10

Problem 10 of 25EasierAlgebra

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?

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Solution

After

m

red-booth and

n

blue-booth exchanges, Alex has

75 - (2m - n)

red tokens,

75-(3n-m)

blue tokens, and

m+n

silver tokens. At termination the red count is

0

1,

and the blue count is

0,1,

2.

Solving the token equations over these six cases leaves only

(m,n)=(59,44),

ending at

(1,2),

(m,n)=(60,45),

ending at

(0,0).

The latter is unreachable: its final exchange would have to start at either

(-1,3)

(2,-1).

Hence Alex finishes with

59+44=103

silver tokens. Thus, the correct answer is E.

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