2013 AMC 10B Problem 17

1. Each trade, at either booth, gives exactly one silver token. So the number of silver tokens equals the number of trades. We do not need to know the order of the trades, only how many of each kind.
2. Let $x$ be the number of trades at the red booth (give $2$ red, get $1$ blue) and $y$ the number at the blue booth (give $3$ blue, get $1$ red). At the end,
$$\text{red} = 75 - 2x + y, \qquad \text{blue} = 75 + x - 3y.$$
3. Trading stops when Alex has fewer than $2$ red and fewer than $3$ blue. So at the end, red is $0$ or $1$ and blue is $0$, $1$ or $2$. That is six possible endings. Which one happens?
4. To decide, find a quantity that changes in a predictable way. Try red $+ 2 \times$ blue:
- Red booth: red goes down $2$, blue goes up $1$, so the quantity changes by $-2 + 2 = 0$.
- Blue booth: blue goes down $3$, red goes up $1$, so the quantity changes by $1 - 6 = -5$.

It starts at $75 + 2 \times 75 = 225$, a multiple of $5$, and it only ever drops by $5$. So it is always a multiple of $5$.
5. Check the six endings:

Only $(0, 0)$ and $(1, 2)$ give multiples of $5$. Every trade hands Alex back a red or blue token, so he can never be left with none. The ending is $1$ red and $2$ blue.
6. The quantity went from $225$ down to $1 + 2 \times 2 = 5$, a drop of $220$. Each blue-booth trade drops it by $5$, so $y = 220 \div 5 = 44$.
7. Use the red count: $75 - 2x + 44 = 1$, so $2x = 118$ and $x = 59$. Check blue: $75 + 59 - 3 \times 44 = 134 - 132 = 2$. Correct.
8. Total trades: $59 + 44 = 103$.

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

Acting out $103$ trades would take forever, but each type of trade changes the counts by the same fixed amounts every time, so only the number of each type matters. A quantity that changes predictably (here red $+ 2\times$blue) tells you where the process must end, and then simple equations give the counts.

Once you know the ending is $1$ red and $2$ blue, just solve the two equations from step 2:
$$75 - 2x + y = 1, \qquad 75 + x - 3y = 2.$$
The first gives $y = 2x - 74$. Put that into the second: $75 + x - 3(2x - 74) = 2$, so $297 - 5x = 2$, $x = 59$, and $y = 118 - 74 = 44$. Again $59 + 44 = 103$.