1992 AIME Problem 15

Let $f(m)=\sum_{j\geq1}\lfloor \frac{m}{5^j}\rfloor$, the number of trailing zeroes in $m!$

Every positive value attained by $f$ first appears at a multiple $5k$, where $f(5k)=k+f(k)$

The number of trailing zeroes is $f(m)=\sum_{j\geq1}\lfloor \frac{m}{5^j}\rfloor.$ Its positive distinct values occur at the multiples $5k,$ and $$f(5k)=k+f(k)$$ is strictly increasing with $k.$ Now $$\begin{aligned}f(1595)&=319+63\\&\quad+12+2=396,\end{aligned}$$ so $f(7975)=1595+396=1991.$ For $k=1596,$ the same calculation gives $f(k)=396,$ hence $f(5k)=1992.$ Therefore exactly $1595$ positive values through $1991$ are factorial tails. Of the $1991$ positive integers below $1992,$ the number omitted is $1991-1595=396.$