year 10 factorial question I would like to know the number of zeros occuring in the factorial of 2016? (2016!)
I have read some ways but i don't understand it.
 A: In order to find the number of trailing zeros you need to determine how many times $10$ divides $2016!$.  In order for a factor of 10 to be present you need a (prime) factor of $2$ and a (prime) factor of $5$.  So think about how many number are divisible from $1, 2, 3, ..., 2016$ are divisible by $5$?  There are $\lfloor \frac{2016}{5}\rfloor$ of them.  But this is not all.  Some numbers are divisible by 5 twice (i.e. multiples of 25).  How many multiples of 25 are there (1 extra prime factor for each multiple)?  There are $\lfloor \frac{2016}{25}\rfloor$ of them.  Some of those numbers are also divisible by 5 three times (i.e. multiples of 125).  Find the number of those in the same way.  Follow this procedure until $5^{k} > 2016$.  Add up all those values and you have the number of times $5$ appears as a factor.  You can similarly count how many times $2$ occurs, but all you need to be certain of is that $2$ occurs at least as many times as $5$ does (I'll leave this to you to determine, not hard).
A: A simple approach to the problem would be as follows:


*

*$5^1$: 2016÷5 = 403.2, so I have 403 factors of 5

*$5^2$: 2016÷25 = 80.64, so I have 80 factors of 25

*$5^3$: 2016÷125 = 16.128, so I have 16 factors of 125

*$5^4$: 2016÷625 = 3.22, so I have 3 factors of 625

*$5^5$: 2016÷3125 < 1, so I stop here.


In total, I now have 403+80+16+3 = 502 trailing zeroes.
