8. There are $(\quad)$ different pairs of digits $(a, b)$ such that $\overline{5 a 68} \times \overline{865 b}$ is divisible by 824. Among them, if $a=0$, then $b=0, b=1, b=2$ all satisfy the requirement, which corresponds to three different pairs.
(A) 10
(B) 11
(C) 15
(D) 19
(E) 21
8. D.

Since $824=2^{3} \times 103$, and 103 is a prime number, at least one of $\overline{5 a 68}$ and $\overline{865 b}$ must be a multiple of 103.

If $865 b$ is a multiple of 103, then $b=2$. In this case, since $\overline{5 a 68}$ is a multiple of 4 and 8652 is a multiple of 2, $\overline{5 a 68} \times 8652$ must be a multiple of 824, and $a$ can take any value from $0 \sim 9$, for a total of 10 cases.

If $\overline{5 a 68}$ is a multiple of 103, it is easy to find that $a=7$. In this case, 5768 is a multiple of 8, and $5768 \times \overline{865 b}$ must be a multiple of 824, and $b$ can take any value from $0 \sim 9$, for a total of 10 cases.

Since $a=7, b=2$ is counted twice, the total number of different pairs $(a, b)$ is $10+10-1=19$.