2023 AMC 10B Problem 11

Organize the count by the number of $\$100$ bills, and for each case find how many ways the rest can be paid.

1. The fifties come in pairs. Twenties and hundreds always add up to a multiple of $\$20$, and $\$800$ is a multiple of $\$20$ too. So the fifties must also add up to a multiple of $\$20$. One fifty is $\$50$, two are $\$100$, three are $\$150$, four are $\$200$: only an even number of fifties works. So the number of fifties is $2, 4, 6, \ldots$ (at least $2$, since there is at least one).
2. One hundred-dollar bill. That leaves $\$700$ for twenties and fifties. Try each even number of fifties:
- $2$ fifties $= \$100$, leaving $\$600 = 30$ twenties.
- $4$ fifties $= \$200$, leaving $\$500 = 25$ twenties.
- $6, 8, 10, 12$ fifties leave $\$400, \$300, \$200, \$100$: $20, 15, 10, 5$ twenties.
- $14$ fifties $= \$700$ leaves no twenties, which is not allowed.

So there are $6$ ways.
3. Each extra hundred removes one way. Two hundreds leave $\$600$: fifties can be $2, 4, \ldots, 10$ (with $12$ fifties the twenties run out), which is $5$ ways. In general, $\$100$ less means one fewer pair of fifties fits.

With $6$ hundreds, $\$200$ is left: $2$ fifties and $5$ twenties is the only way. With $7$ hundreds, $\$100$ is left, and $2$ fifties would use it all with no twenties, so there is no way.
4. Add up. $6 + 5 + 4 + 3 + 2 + 1 = 21$.

The answer is $\boxed{\textbf{(B)}\ 21}$.

Looking at what each bill adds up to shows that the fifties must come in pairs, which cuts the choices in half. After that, fixing one kind of bill at a time turns the problem into short lists you can count. Build the "at least one of each" rule into every case from the start, so you do not have to remove extra cases at the end.

Use letters. Let $a$, $b$, $c$ be the numbers of $\$20$, $\$50$, $\$100$ bills, each at least $1$. Dividing $20a + 50b + 100c = 800$ by $10$ gives $2a + 5b + 10c = 80$.

1. $2a$, $10c$ and $80$ are even, so $5b$ is even, so $b$ is even. Write $b = 2k$ with $k \ge 1$.
2. Then $2a + 10k + 10c = 80$, so $a = 40 - 5(k + c)$.
3. $a$ must be at least $1$, so $k + c$ is at most $7$. Also $k$ and $c$ are at least $1$, so $k + c$ is at least $2$.
4. If $k + c = s$, there are $s - 1$ choices for $(k, c)$, and $a$ is then fixed. Adding over $s = 2, 3, \ldots, 7$: $1 + 2 + 3 + 4 + 5 + 6 = 21$.