$33 \cdot 16$ Let $S=1!+2!+3!+\cdots+99!$, then the units digit of $S$ is
(A) 9.
(B) 8.
(C) 5.
(D) 3.
(E) 0.
(19th American High School Mathematics Examination, 1968)
［Solution］Notice that $k!$ for $k \geqslant 5$ all have factors of 2 and 5, so their product is a multiple of 10.

Thus, the units digit of $S$ is the same as that of $1!+2!+3!+4!$, which is the units digit of $(1+2+6+24)$, which is 3.
Therefore, the answer is $(D)$.