# 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.

• Number of trailing zeros or the total number of them? Because the second one is way harder. – rubik Jul 27 '16 at 15:10
• According to the PARI/GP program ? q=0;x=digits(2016!);for(j=1,length(x),if(x[j]==0,q=q+1));print(q) 1006 ? , $2016!$ contains $1006$ zeros. The number of trailing zeros is $502$. – Peter Jul 27 '16 at 15:11
• total number. No wonder...the one i searched about were the solutions to the trailing zeros – 1234 Jul 27 '16 at 15:11
• The easiest way to find the number of trailing zeros : Divide $2016$ repeatedly by $5$ ignoring the remainder until you arrive at $0$ and add all occuring numbers (except $2016$, of course). You get the sequence $403,80,16,3,0$ , which sums up to $502$. – Peter Jul 27 '16 at 15:19
• The idea is quite simple : You have to count how many numbers from $1$ to $n$ are divisble by $5$, by $5^2$ , by $5^3$ and so on until we reach a power of $5$ greater than $n$ (here, the number is $0$). There are $n_1=trunc(n/5)$ numbers from $1$ to $n$ divisble by $5$, so we start with $n_1=trunc(n/5)$. $n_2=trunc(n_1/5)$ of these numbers are divisble by $5^2$, the exponent in $5^m$ will increase by $n_2$ because we already covered exponent $1$ with $n_1$. You can continue until $n_k=0$ and the sum of the $n_i$'s is the desired number. – Peter Jul 27 '16 at 16:14

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).

• There may be a nicer "number theory" way to do this. This is a "combinatorial" way to solve it. – TravisJ Jul 27 '16 at 16:14

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.