$29 \cdot 25$ Determine the largest positive integer $n$ such that $1005!$ is divisible by $10^n$.
(A) 102.
(B) 112.
(C) 249.
(D) 502.
(E) None of these.
(28th American High School Mathematics Examination, 1977)
[Solution] If $2^{k} 3^{l} 5^{m} \cdots$ is the factorization of 1005! into powers of distinct primes, then $n=\min \{k, m\}$.

Now there are 201 integers between 5 and 1005 that are divisible by 5; among these 201 integers, 40 are divisible by $5^{2}$; among these 40 integers, 8 are divisible by $5^{3}$; among these 8 integers, 1 is divisible by $5^{4}$.
Thus, $m=201+40+8+1=250$.
Also,
$$
k>\left[\frac{1005}{2}\right]=502 \text {, }
$$

Therefore,
$$
n=\min \{k, m\}=250 \text {. }
$$

Hence, the answer is $(E)$.