2013 AMC 10B Problem 17

Every exchange produces one silver token, so we need the total number of exchanges. Let $x$ be the number of red-booth exchanges (red $-2$, blue $+1$) and $y$ the number of blue-booth exchanges (blue $-3$, red $+1$). At the end,
$$\text{red} = 75 - 2x + y, \qquad \text{blue} = 75 + x - 3y.$$

Which final state is possible? Exchanges stop only when red $\le 1$ and blue $\le 2$. The quantity $R + 2B$ is unchanged by a red exchange ($-2 + 2$) and drops by $5$ under a blue exchange, so it stays a multiple of $5$ (it starts at $225$). Among the six candidate end states, only $(0, 0)$ and $(1, 2)$ have $R + 2B$ divisible by $5$, and $(0, 0)$ is unreachable since every exchange hands back a token. So Alex ends with $1$ red and $2$ blue.

Solve:
$$75 - 2x + y = 1, \qquad 75 + x - 3y = 2.$$
From the first, $y = 2x - 74$; substituting, $75 + x - 6x + 222 = 2$, so $5x = 295$, $x = 59$, $y = 44$.

Total exchanges: $x + y = 103$.

The answer is $\boxed{\textbf{(E)}\ 103}$.

Simulating $103$ steps is hopeless, but the order of exchanges does not matter for the final count: each type of exchange changes the token counts by a fixed vector, so the end state is a linear function of how many of each were done. Identify the terminal state (here with a small invariant), then solve the linear system.

Use two invariants instead of a system. $R + 2B$ falls by $5$ per blue exchange only: $225 - 5y = 1 + 4$, so $y = 44$. $R + B$ falls by $1$ per red exchange and $2$ per blue: $150 - x - 2y = 3$, so $x = 59$. Again $x + y = 103$.

• Trying to simulate all the exchanges by hand, or assuming the process ends with 0 tokens of each color.
• Counting only one booth's exchanges (e.g. $y = 44$ plus a miscount) and landing on choices (B) or (C).