2020 AMC 10A Problem 6

Count digit by digit, starting with the digit that has the strictest rule.

1. Last digit. A number is divisible by $5$ when its last digit is $0$ or $5$. The digit must also be even, and $5$ is odd. So the last digit must be $0$: $1$ choice.
2. First digit. It must be even, and a $4$-digit number cannot start with $0$. So it is $2$, $4$, $6$ or $8$: $4$ choices.
3. Second and third digits. Each can be any even digit $0, 2, 4, 6, 8$: $5$ choices each.
4. The choices do not affect each other, so multiply:
$$4 \times 5 \times 5 \times 1 = 100.$$

For example, $2000$, $2020$ and $8680$ are all counted.

The answer is $\boxed{\textbf{(B)}\ 100}$.

When each digit's choices do not depend on the others, the total count is the product of the number of choices for each digit. Settle the most restricted spots first (here the last digit is forced), and always remember that the first digit of a number cannot be $0$.

A number is divisible by $5$ exactly when its last digit is $0$ or $5$. Since every digit must be even, the last digit is forced to be $0$: one choice.

The thousands digit must be even and nonzero: $2, 4, 6, 8$, four choices.

The hundreds and tens digits may be any even digit $0, 2, 4, 6, 8$: five choices each.

By the multiplication principle the count is
$$4 \cdot 5 \cdot 5 \cdot 1 = 100.$$

The answer is $\boxed{\textbf{(B)}\ 100}$.