2013 AMC 10B Problem 17

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. Starting from $(75,75)$ red and blue tokens, make $25$ blue-booth exchanges, then $50$ red-booth exchanges, then $16$ blue-booth exchanges, then $8$ red-booth exchanges, then $3$ blue-booth exchanges, and finally $1$ red-booth exchange. The red-blue counts become $$\begin{aligned} &(100,0),(0,50),(16,2),\\ &(0,10),(3,1),(1,2). \end{aligned}$$ Therefore Alex ends with $59+44=103$ silver tokens, and the correct answer is E .