2023 AMC 10B Problem 11

Let $a, b, c \ge 1$ be the numbers of $\$20$, $\$50$ and $\$100$ bills. Then $20a + 50b + 100c = 800$, or after dividing by $10$,
$$2a + 5b + 10c = 80.$$
Every term except $5b$ is even, so $b$ is even: write $b = 2k$ with $k \ge 1$. Dividing by $2$,
$$a + 5k + 5c = 40, \qquad a = 40 - 5(k + c).$$
Since $a \ge 1$, we need $k + c \le 7$; since $k, c \ge 1$, we need $k + c \ge 2$. Each pair $(k, c)$ with $k + c = s$ determines $a$, and there are $s - 1$ such pairs. Summing over $s = 2, 3, \dots, 7$:
$$1 + 2 + 3 + 4 + 5 + 6 = 21.$$

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

Divisibility shrinks the search: the $\$50$ bills must come in pairs because everything else is a multiple of $20$. After the substitution the equation has a single free combination $k + c$, and counting positive pairs with a given sum is routine. Handle the "at least one" restriction by building it into the variable ranges from the start.

Casework on $c$: for $c = 1, \dots, 6$ the remaining $800 - 100c$ dollars must be made from $a \ge 1$ twenties and an even number $b \ge 2$ of fifties, giving $6, 5, 4, 3, 2, 1$ options respectively, again $21$.

• Ignoring 'at least one of each denomination', which allows zero counts and inflates the total to 45.
• Forgetting that $b$ must be even and counting odd values of $b$ that leave an odd number of tens.