Exponent of prime $p$ in $n!$

Question

For how many positive integral values of $n$ does $n!$ end with precisely $25$ zeroes?

My work

Number of zeroes at the end of $n!$=$25$

Using Legendre's Theorem,

$$E_5(n!)=25$$

$$\lfloor \frac{n}{5} \rfloor+\lfloor \frac{n}{25} \rfloor +...=25.$$

How to solve this? Can you give me hint?

Answer 1

You solved upto this step.

$$\lfloor \frac{n}{5} \rfloor+\lfloor \frac{n}{25}\rfloor+...=25$$

First put $n$=$100$.

$$\lfloor \frac{100}{5} \rfloor+\lfloor \frac{100}{25}\rfloor+...=20+4=24$$

You need one more $0$.

Now put $n$=$105$

$$\lfloor \frac{105}{5} \rfloor+\lfloor \frac{105}{25}\rfloor+...=21+4=25$$

This continues upto $n$=$109$.

So, there are 5 such integers.