2000 AMC 10 Problem 17

1. Watch what each trade does to the total amount of money, in cents.
- A quarter ($25$ cents) becomes five nickels ($25$ cents). No change.
- A nickel ($5$ cents) becomes five pennies ($5$ cents). No change.
- A penny ($1$ cent) becomes five quarters ($125$ cents). A gain of $124$ cents.
2. Boris starts with $1$ cent. Only the penny trade changes the total, and each time it adds $124$. So his total is always
$$1,\ 1 + 124,\ 1 + 2 \cdot 124,\ \ldots$$
that is, $1$ more than a multiple of $124$.
3. Subtract $1$ from each choice and see which result divides evenly by $124$:

1. Only $\$7.45$ works. It really can happen: he always has a penny available (break a quarter into nickels and a nickel into pennies if needed), so he can do the penny trade $6$ times.

The answer is $\boxed{\textbf{(D)}\ \$7.45}$.

When a process has only a few kinds of moves, ask what each move does to one simple number, here the total in cents. Two moves leave it alone and one adds a fixed $124$, so the total is always "start plus some number of 124s." That one fact rules out four choices without acting out a single trade.

Track the total value in cents through each kind of exchange.

• Quarter in, five nickels out: $25 \to 25$, no change.
• Nickel in, five pennies out: $5 \to 5$, no change.
• Penny in, five quarters out: $1 \to 125$, a gain of $124$.

So no matter what Boris does, his total is $1 + 124k$ cents, where $k$ counts how many pennies he has fed in. Subtract $1$ from each choice and test divisibility by $124$:
$$362 = 2 \cdot 124 + 114,\quad 512 = 4 \cdot 124 + 16,\quad 629 = 5 \cdot 124 + 9,\quad 744 = 6 \cdot 124,\quad 906 = 7 \cdot 124 + 38.$$
Only $\$7.45$ fits, with $k = 6$. (It is also reachable: each time, break a quarter into nickels and a nickel into pennies, then feed in a penny.)

The answer is $\boxed{\textbf{(D)}\ \$7.45}$.