Proof that $n=250$ is the largest value for $n$ for which $1005!$ is divisible by $10^n$ where $n\in\Bbb{N}$

Question

Proof that $n=250$ is the largest value for $n$ for which $1005!$ is divisible by $10^n$ where $n\in\Bbb{N}$.

My attempt was to find a way for counting the number of last zeroes for any permutation $k!$ where $k\in\Bbb{N}$, however I haven't arrived at any useful result. I don't know how to proceed with this one.

Answer 1

If $10^n\mid 1005!$, then $2^n\mid 1005!$ and $5^n\mid 1005!$. Hence, it suffices to find the largest $n$ such that $5^n\mid 1005!$. To count how many factors of $5$ will go into $1005!$, you need to count how many numbers are divisible by $5$ that are less than or equal to $1005$. This is $\lfloor\frac{1005}{5}\rfloor$. But this doesn't account for a number you multiply by in the factorial which is not only divisible by $5$, but by $5^2$. You can count these by looking at $\lfloor\frac{1005}{5^2}\rfloor$. Similarly, you can see how many numbers are not only divisible by $5^2$ but by $5^3$, $5^4$, and so on. Hence, you need to compute $$ \sum_{i = 1}^{\infty}\lfloor\frac{1005}{5^i}\rfloor. $$ (Eventually, $5^i$ is bigger than $1005$, so all terms after that will be $0$, and this is really a finite sum.)

Answer 2

The question is equivalent to when we consider the prime factorization of $1005!$, what is the index for base $5$? Do you see why?

We just need to compute

$$\lfloor \frac{1005}{5} \rfloor+\lfloor \frac{1005}{5^2} \rfloor+\lfloor \frac{1005}{5^3} \rfloor+\lfloor \frac{1005}{5^4} \rfloor=250$$