Find number of zeros at the end of $2007!$ using greatest integer function?

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How to solve this question using greatest integer function? I used logical thinking to get answer but I want to solve this question using greatest integer function.

Answer: $500$

Please explain in detail

Please don't close my question. Since I am just a high school student seeking help from teachers.

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You can easily justify that there are more factors of $2$ in $2007!$ than there are factors of $5$. So you get a $0$ for every number divisible by $5$, like $5,10,15,20,...$. It is easy to see that there are $401$ such numbers ($5\cdot1,5\cdot2,...,5\cdot401$). But all multiples of $25$ are counted only once, when they contain at least an extra factor of $5$. There are $80$ such multiples, so we add $80$ extra zeros. Then one notice that $125$ and multiples contribute an extra zero each. And then you need to repeat the procedure for $625=5\cdot5\cdot5\cdot5$. You will end up with $401+80+16+3=500$. Note that $625\cdot5=3125$ is greater than $2007$, so you don't need to worry about that.