8. Given a six-digit decimal number composed of six positive integers, the digit in the units place is a multiple of 4, the digits in the tens and hundreds places are multiples of 3, and the sum of the digits of the six-digit number is 21. Then the number of six-digit numbers that satisfy the above conditions is $\qquad$
8. 126 .

It is easy to know that the last three digits can only be:
$$
334,338,664,364,368,394,634,638 \text {, }
$$
934.
Also, the sum of the first three digits of 334 is 11, so there are $\mathrm{C}_{10}^{2}$ ways.
Thus, $\mathrm{C}_{10}^{2}+\mathrm{C}_{6}^{2}+\mathrm{C}_{4}^{2}+2\left(\mathrm{C}_{7}^{2}+\mathrm{C}_{3}^{2}+\mathrm{C}_{4}^{2}\right)$
$$
\begin{array}{l}
=45+15+6+2(21+3+6) \\
=66+2 \times 30=126 .
\end{array}
$$