16. Determine the number of 0 's at the end of the value of the product $1 \times 2 \times 3 \times 4 \times \cdots \times 2008$.
16. Answer: 500 .
Let $m$ be the maximum integer such that $2^{m}$ is a factor of $1 \times 2 \times 3 \times 4 \times \cdots \times 2008$, and $n$ the maximum integer such that $5^{n}$ is a factor of $1 \times 2 \times 3 \times 4 \times \cdots \times 2008$.
Then the number of 0 's at the end of $1 \times 2 \times 3 \times 4 \times \cdots \times 2008$ is equal to in $\{m, n\}$. It is obvious that $m \geq n$.
Observe that $2008<5^{5}$. For $1 \leq k \leq 4$, the number, denoted by $a_{k}$, of integers in $\{1,2.3,4, \cdots, 2008\}$, denoted by $\bar{A}$, divisible by $5^{k}$ is
$$
a_{k}=\left\lfloor 2008 / 5^{k}\right\rfloor \text {. }
$$

So $a_{1}=401, a_{2}=80, a_{3}=16, a_{4}=3$. Thus, there are exactly 3 numbers in $A$ divisible by $5^{4}$;
exactly $16-3=13$ numbers in $A$ divisible by $5^{3}$ but not by $5^{4}$;
exactly $80-16=64$ numbers in $A$ divisible by $5^{2}$ but not by $5^{3}$;
exactly $401-80=321$ numbers in $A$ divisible by 5 but not by $5^{2}$.
Hence, the maximum integer $n$ such that $5^{n}$ is a factor of $1 \times 2 \times 3 \times 4 \times \cdots \times 2008$ is
$$
n=4 \times 3+3 \times 13+2 \times 64+1 \times 321=500 .
$$