1992 AIME Problems/Problem 15

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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 positiive integers less than $1992$ are not factorial tails?

Solution

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

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 $1992$ than are not factorial tails.

1992 AIME (Problems • Resources)
Preceded by Problem 14	Followed by Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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1992 AIME Problems/Problem 15

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