15. If there exists a positive integer $m$ such that $m!$ ends with exactly $n$ zeros, then the positive integer $n$ is called a "factorial tail number." How many non-"factorial tail number" positive integers are there less than 1992?
15. Let $f(m)$ be the number of trailing zeros in the number $m!$. Clearly, $f(m)$ is a non-decreasing function of $m$, and when $m$ is a multiple of 5, we have
$$
\begin{aligned}
f(m) & =f(m+1)=f(m+2) \\
& =f(m+3)=f(m+4) \\
& 4 \times 1991=7964. \text{ Using (2), it is easy to get }
$$
$$
f(7965)=1988, f(7975)=1991 .
$$

If we take the sequence (1) from the first term to $f(7979) = 1991$, we get
$$
\begin{array}{l}
\quad 0,0,0,0,0,1,1,1,1,1, \cdots, 1989, \\
1991,1991,1991,1991,1991 .
\end{array}
$$

Sequence (3) contains a total of 7980 numbers. Since each different integer appears exactly 5 times, there are 7980 / 5 = 1596 different integers in sequence (3), and they are all elements of the set $\{0,1, 2, \ldots, 1991\}$. Therefore, there are 1992 - 1596 = 396 integers that do not appear in (3). Thus, there are 396 positive integers less than 1992 that are not "factorial trailing zero numbers". (Solved by Liu Zhi, a student at Peking University)