Question

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?

$62$

$82$

$83$

$102$

$103$

Solution:

Suppose Alex makes $m$ exchanges at the red-token booth and $n$ exchanges at the blue-token booth.

He then has $75-2m+n$ red tokens and $75+m-3n$ blue tokens. At the end he must have fewer than $2$ red tokens and fewer than $3$ blue tokens.

Solving these terminal possibilities gives only two candidate final token counts: $(1,2)$ , which comes from $(m,n)=(59,44)$ , or $(0,0)$ , which comes from $(m,n)=(60,45)$ .

The final count $(0,0)$ is impossible, because the last exchange would always create either one blue token or one red token. The final count $(1,2)$ is attainable, for example by the exchange sequence described in the official construction.

Therefore Alex ends with $59+44=103$ silver tokens, and the correct answer is E .

2013 AMC 10B Problem 17

Solution:

Problem 17 in Other Years