Question

A natural number $n$ such that $n!$ ends in exactly 1000 zeros is

• A. $4010$
• B. $4000$
• C. $4009$
• D. $4004$

Hint:

Trailing zeros in $n!$ are determined by the number of factors of 5.

Answer 1

The Correct Option is C

To count trailing zeros in $n!$, use:
$\left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \cdots$
We want the smallest $n$ for which sum equals 1000. Try $n = 4009$: $\left\lfloor \frac{4009}{5} \right\rfloor = 801$
$\left\lfloor \frac{4009}{25} \right\rfloor = 160$
$\left\lfloor \frac{4009}{125} \right\rfloor = 32$
$\left\lfloor \frac{4009}{625} \right\rfloor = 6$
$\left\lfloor \frac{4009}{3125} \right\rfloor = 1$
Sum = $801 + 160 + 32 + 6 + 1 = 1000$
So, $n = 4009$