Reducing Factorials

The problem is asking, whats the biggest power of 10 that divides $2018!$

If we want to to know the answer, we need to find the number of factors of $2$ and $5$ in $2018!$ because $10 = 2 * 5$ .

We can se that every multiple of 2 add a factor each, as 4 add one more factor (not two because one was alredy count as a multiple of 2), and 8,16, ...

The same happens with 5, 25, 125, 625, etc.

So the answer is $min( \lfloor \frac{2018}{2}\rfloor + \lfloor \frac{2018}{4}\rfloor+...+ \lfloor \frac{2018}{1024}\rfloor,\ \lfloor \frac{2018}{5}\rfloor + \lfloor \frac{2018}{25}\rfloor+\lfloor \frac{2018}{125}\rfloor+ \lfloor \frac{2018}{625}\rfloor ) = min ( 2011, 502 )$ , so 502 is the answer

First, we need to figure out how many factors of 2018! are multiples of 5 since there are always plenty of 2's to multiply with the 5's to make 10. We get this formula: $\lfloor \frac{2018}{5}\rfloor = 403$

However, that is not the end. Powers of 5, such as 25 125 need to be accounted for. Therefore, we need to add up how many times each of the powers of 5 go into 2018:

$\displaystyle\sum_{n=1}^{\infty} \lfloor \frac{2018}{5^n}\rfloor = \boxed{502}$

In general, the amount of trailing zeroes that $n!$ contains can be represented as

$z(n)=\displaystyle\sum_{i=1}^{\infty} \lfloor \frac{n}{5^i}\rfloor$