2022 AMC 10B Problem 3

The last digit can be anything from $0$ to $9$, so it can always "fix" the count. Use it as a switch.

1. Count the first two digits. The hundreds digit is $1$ to $9$ ($9$ choices) and the tens digit is $0$ to $9$ ($10$ choices). That is $9 \cdot 10 = 90$ ways.
2. Look at an example. Say the first two digits are $4$ and $7$. So far there is one even digit (the $4$), which is odd. To keep the count odd, the last digit must be odd: $1, 3, 5, 7, 9$. That is $5$ choices.
3. Another example. Say the first two digits are $4$ and $6$. So far there are two even digits, which is even. To make the count odd, the last digit must be even: $0, 2, 4, 6, 8$. Again $5$ choices.
4. Always 5. In every case, either the $5$ odd last digits work or the $5$ even last digits work. So each of the $90$ starts gives exactly $5$ good numbers.
5. Total. $90 \cdot 5 = 450$.

This is exactly half of the $900$ three-digit numbers, which makes sense: the last digit splits them evenly.

The answer is $\boxed{\textbf{(D)}\ 450}$.

When a question asks whether a count is odd or even, a digit that is completely free (here the last one, which can be any of $0$ to $9$) can flip the answer either way. Half its choices give "odd" and half give "even", so exactly half of all the numbers work. Look for a free position like this before starting long casework.

Count directly. The number of even digits must be $1$ or $3$. Remember the hundreds digit has only $4$ even choices ($2, 4, 6, 8$) and $5$ odd choices, while the other digits have $5$ of each.

• One even digit, and it is the hundreds digit: $4 \cdot 5 \cdot 5 = 100$.
• One even digit, and it is the tens or the units digit: the hundreds digit is odd ($5$ ways), pick which of the other two is even ($2$ ways), then $5$ choices for each: $5 \cdot 2 \cdot 5 \cdot 5 = 250$.
• Three even digits: $4 \cdot 5 \cdot 5 = 100$.

Total: $100 + 250 + 100 = 450$.