Possible values of $n$ such that $n!$ ends with $30$ zeroes

Question

If $n!$ ends with $30$ zeroes, how many values of $n$ are possible?

I know that $10$ will occur when pairs of $2$ and $5$ will occur and since $2$ occurs more times than $5$ does in $n!$, the number of trailing zeroes will be the highest power of $5$.

So I need to find those $n$ for which $n!$ will have $5$ occur $30$ times in its prime factorization.

I can't understand how should I proceed.

I know it has been asked before but I couldn't understand.

Please help.

Answer 1

How many zeros at the end of $0!$?

At the end of $5!$?

Of $10!$?

Of $15!$?

Of $20!$?

Of $25!$?

Go that far and try to understand the pattern, and then extend it to reaching 30 zeros.