Question

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?

Answer

Answer: 396

Concepts:trailing zeros factorial complementary counting

Difficulty rating: 2650

Small Hint:

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

Big Hint:

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

Solution:

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 $1991-1595=396.$

1992 AIME Problem 15

Answer: 396

Difficulty rating: 2650

Small Hint:

Big Hint:

Solution:

Problem 15 in Other Years