Smallest value of n such that the product $n!$ ends in at least 10 zeros.

Question

What is the smallest value of $n$ such that the product $n!$ ends in at least 10 zeros?

I tried to do this by multiplying each number but it didn't work. Please help.

Answer 1

Hint: $10$ consists of the factors $2$ and $5$. How often do those factors arise as you increase the value of $n$? How many of them do you need to get a number with 10 zeros at the end (meaning it is divisible by $10^{10}$)?