I ran into a question in my book, Find $n$ such that $n!$ ends in $12$ zeroes. This can be done with guessing a value of $n$ and improving estimates at each stage until finally being able to judge the exact values of $n$.
They, however, start in the book with the proposition that if $n!$ ends in $\epsilon$ zeroes then $n\geq 4\epsilon$.
Instead of the proof, I'd like to know why it's obvious and why it makes sense. I'm comfortable with the basics of combinatorics if that'll come into use. Thanks
In the factorization of $n!$, $5$'s are less plentiful then $2$'s, so the number of zeroes is just the number of $5$'s. There are $\lfloor n/5\rfloor$ multiples of $5$ which are less than $n$. In addition, the $\lfloor n/25\rfloor$ multiples of $25$ in $n!$ contribute an extra $5$ each. Same for the multiples of $5^3=125$, and of $5^4=625$, and so on. Therefore, \begin{align} \text{# zeroes} &=\lfloor n/5\rfloor+\lfloor n/5^2\rfloor+\lfloor n/5^3\rfloor+\dots \\&\le n/5+n/5^2+n/5^3+\dots \\&=\frac{n/5}{1-1/5} \\&=n/4. \end{align} Therefore, if $\text{# zeroes}=\epsilon$, we have $n\ge 4\epsilon$.