$21 、$ A three-digit number is a multiple of 6, and the sum of the units digit and the hundreds digit is twice the tens digit, how many such three-digit numbers are there $\qquad$?
Let's analyze: Suppose this three-digit number is $\overline{a b c}$. Since $\overline{a b c}$ is a multiple of 6, $c$ must be even. Also, since $a+c=2b$, $a$ must also be even. The sum of the digits $a+b+c=3b$ is definitely a multiple of 3, so it is sufficient to ensure that both $a$ and $c$ are even. The possible values for $a$ are $2, 4, 6, 8$, and the possible values for $c$ are $0, 2, 4, 6, 8$. By the principle of multiplication, we get: $4 \times 5=20$ (numbers)