14.54 If there is a positive integer $m$, such that $m$! ends with exactly $n$ zeros, then the positive integer $n$ is called a "factorial tail". How many positive integers less than 1992 are not "factorial tails"?
(10th American Mathematical Invitational, 1992)
[Solution] Let $f(m)$ be the number of trailing zeros in the number $m!$.
Obviously, $f(m)$ is a non-decreasing function of $m$.
Moreover, when $m$ is a multiple of 5, we have
$$
\begin{aligned}
f(m) & =f(m+1)=f(m+2)=f(m+3) \\
& =f(m+4)4 \cdot 1991=7964$.
Using (2), it is easy to get
$$
f(7965)=1988, \quad f(7970)=1989, \quad f(7975)=1991 .
$$

Now, we list the sequence (1) up to $f(7979)=1991$, obtaining
$$
0,0,0,0,0,1,1,1,1,1, \cdots, 1989,1989,1989,1989,1989,1991 \text {, }
$$
$1991,1991,1991,1991$.
Sequence (3) contains a total of 7980 numbers.
Since each different integer appears exactly 5 times, sequence (3) contains $\frac{7980}{5}=$ 1596 different integers, all of which are elements of the set $\{0,1,2, \cdots, 1991\}$.
Therefore, there are $1992-1596=396$ integers that do not appear in sequence (3).
Thus, there are 396 positive integers less than 1992 that are not "factorial tail numbers".