1992 AIME Problems/Problem 15

Problem

Define a positive integer $n^{}_{}$ to be a factorial tail if there is some positive integer $m^{}_{}$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails?

Solution

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

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 $f(m)$ 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 $319 + 63 + 12 + 2 = \boxed{396}$ .

1992 AIME (Problems • Answer Key • Resources)
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1992 AIME Problems/Problem 15

Contents

Problem

Solution

Solution 2

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