AIME 1992 · 第 15 题

Solution

Let the number of zeros at the end of $m!$ be $f(m)$. We have $f(m) = \left\lfloor \frac{m}{5} \right\rfloor + \left\lfloor \frac{m}{25} \right\rfloor + \left\lfloor \frac{m}{125} \right\rfloor + \left\lfloor \frac{m}{625} \right\rfloor + \left\lfloor \frac{m}{3125} \right\rfloor + \cdots$.

Note that if $m$ is a multiple of $5$, $f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)$.

Since $f(m) \le \frac{m}{5} + \frac{m}{25} + \frac{m}{125} + \cdots = \frac{m}{4}$, a value of $m$ such that $f(m) = 1991$ is greater than $7964$. Testing values greater than this yields $f(7975)=1991$.

There are $\frac{7975}{5} = 1595$ distinct positive integers, $f(m)$, less than $1992$. Thus, there are $1991-1595 = \boxed{396}$ positive integers less than $1992$ that are not factorial tails.

Solution 2

After testing various values of $m$ in $f(m)$ of solution 1 to determine $m$ for which $f(m) = 1992$, we find that $m \in \{7980, 7981, 7982, 7983, 7984\}$. WLOG, we select $7980$. Furthermore, note that every time $k$ reaches a multiple of $25$, $k!$ will gain two or more additional factors of $5$ and will thus skip one or more numbers.

With this logic, we realize that the desired quantity is simply $\left \lfloor \frac{7980}{25} \right \rfloor + \left \lfloor \frac{7980}{125} \right \rfloor \cdots$, where the first term accounts for every time $1$ number is skipped, the second term accounts for each time $2$ numbers are skipped, and so on. Evaluating this gives us $319 + 63 + 12 + 2 = \boxed{396}$. - Spacesam(edited by srisainandan6)