16. A seven-digit number $\overline{m 0 A 0 B 9 C}$ is a multiple of 33. We denote the number of such seven-digit numbers as $a_{m}$. For example, $a_{5}$ represents the number of seven-digit numbers of the form $\overline{50 A 0 B 9 C}$ that are multiples of $\mathbf{3 3}$. Then $a_{2}-a_{3}=$
【Analysis】$\overline{m 0 A 0 B 9 C}$ is a multiple of 33, i.e., $m+A+B+\overline{9 C}=90+m+A+B+C$ is a multiple of 33 when $m=2$, $92+A+B+C$ is a multiple of 33, since $92+A+B+C \leq 92+27=119$, so $92+A+B+C=99, A+B+C=7$, i.e., $(A+1)+(B+1)+(C+1)=10$, by the stars and bars method, there are $C_{9}^{2}=36$ numbers that meet the requirement, i.e., $a_{2}=36$

When $m=3$, $93+A+B+C$ is a multiple of 33, since $93+A+B+C \leq 93+27=120$, so $93+A+B+C=99, A+B+C=6$, i.e., $(A+1)+(B+1)+(C+1)=9$, by the stars and bars method, there are $C_{8}^{2}=28$ numbers that meet the requirement, i.e., $a_{3}=28$ thus $a_{2}-a_{3}=8$