# A Note on Factorial and it's Trailing Zeros

Published by Ganit Charcha | Category - Math Articles | 2014-10-06 08:09:28

Definition
For any positive integer $n (>0)$ the factorial of $n$, denoted as $n!$, is defined as
\begin{align}
n! = n.(n-1).(n-2)\ldots.2.1. \hspace{2.2in}     
\end{align}
$0!$ is defined as $1$, i.e., $0! = 1,$ though it looks bit unintutive. From the definition, it is clear that
\begin{align}
n! = n.(n-1)! \hspace{3.5in}     
\end{align}
Now, for $n = 1$, we have $1! = 1.0!$ and this implies that $0! = 1$. So, now we see that the definition of $0!$ is in accordance with the equation (2). The figure below shows the values of the factorial for the integers $1$ to $16$ and we can see the valus of the factorial are getting bigger quickly as $n$ grows from $1$ to $16$. Also note that, for $n > 4$ each factorail ends with one or more $0$'s at the end. Counting the Trailing Zeros
Indeed interesting question - how do we find out the number of trailing zeros in a factorial. The task looks really daunting when the number whose factorial is to be computed itself is big. The list above shows that the one trailing zero appeared for the first time in $5!$ and the same continued for $6!$, $7!$, $8!$ and $9!$ also. The number of trailing zeros becomes $2$ in $10!$ and the same continued till $14!$. In $15!$ number of trailing zeros becomes $3$. Do we see any pattern now? Of course we see and we are really in hurry now to conclude that $20!$ to $24!$ will have $4$ trailing zeros, $25!$ to $29!$ will have $5$ trailing zeros, $30!$ to $34!$ will have $6$ trailing zeros and so on.

Let us now take a look at, $$24! = 620,448,401,733,239,439,360,000$$ and $$25! = 15,511,210,043,330,985,984,000,000$$ We are now surprised to have seen $6$ trailing zeros instead of $5$ in $25!$. What went wrong?
The number $5$ when multiplied by any even number contributes a trailing zero at the end of the factorial. Up to $4!$, no number is divisible by $5,$ so there is no trailing zero.

(1) From $5!$ to $9!$, there is only one number,i.e. $5$ which is divisible by $5$, so there is only one trailing zero.
(2) From $10!$ to $14!$, there are two numbers $5$ and $10$ which are divisible by $5$, so there are $2$ trailing zeros.
(3) From $15!$ to $19!$, there are three numbers $5$, $10$ and $15$ which are divisible by $5$, so there are $3$ trailing zeros.
(4) From $20!$ to $24!$, there are four numbers $5$, $10$, $15$ and $20$ which are divisible by $5$, so there are $4$ trailing zeros.

In $25!,$ there are $5$ numbers $5$, $10$, $15$, $20$ and $25$ that are divisible by $5,$ but the number $25$ itself contributes one additional $5$ as a factor. This additional factor will also contribute one more trailing zero. That is why $25!$ has $6$ trailing zeros.

$n!$ when will be expressed as products of powers of primes will take the form
\begin{align}
\label{eq:3}
n! = 2^{c_1}.5^{c_2}.p_{1}^{c_3}.p_{2}^{c_4}.p_{3}^{c_5}\ldots, \hspace{3in}     
\end{align}
where $p_{i}$ ($i = 1,2,3,\ldots$) are primes other than $2$ and $5$ and $c_{i}$'s for $i = 1, 2, 3, \ldots$ are constants greater than $0$.
In $n!$, there are $\lfloor \frac{n}{2} \rfloor$ number of integers which are even and there are $\lfloor \frac{n}{5} \rfloor$ number of integers which are divisible by $5$ and $\lfloor \frac{n}{2} \rfloor > \lfloor \frac{n}{5} \rfloor.$ Also, if $m$ is the highest power of $5$ such that $k.5^{m} < n$ ($k > 0$), then $k.2^{m} < n$. In fact we have the following,
\begin{align}
\label{eq:4}
\lfloor \frac{n}{2^{j}} \rfloor > \lfloor \frac{n}{5^{j}} \rfloor \hspace{0.1in} for \hspace{0.1in} j = 1,2,\ldots,m \hspace{2.5in}     
\end{align}
and also
\begin{align}
\label{eq:5}
\lfloor\frac{n}{2^{j+1}} \rfloor > \lfloor \frac{n}{5^{j}}\rfloor \hspace{0.1in} for \hspace{0.1in} j = 1,2,\ldots,m \hspace{2.3in}     
\end{align}
\begin{align}
\label{eq:6}
=> \lfloor \frac{n}{2^{j}}\rfloor > 2.\lfloor \frac{n}{5^{j}}\rfloor \hspace{0.1in} for \hspace{0.1in} j = 1,2,\ldots,m \hspace{2.0in}     
\end{align}
Hence, for $j = 1, 2, \ldots, m,$ we have
\begin{align}
\label{eq:7}
\lfloor \frac{n}{2^{j}}\rfloor - \lfloor \frac{n}{2^{j+1}}\rfloor > \lfloor \frac{n}{5^{j}}\rfloor - \lfloor \frac{n}{5^{j+1}}\rfloor \hspace{2.5in}     
\end{align}
and also we have
\begin{align}
\label{eq:8}
\lfloor \frac{n}{2^{j}}\rfloor - \lfloor \frac{n}{2^{j+1}}\rfloor > \lfloor \frac{n}{5^{j}}\rfloor \hspace{3.0in}     
\end{align}
Denoting $\lfloor \frac{n}{5^{j}}\rfloor = c_{2j}$ for $j = 1, 2, \ldots, m$, $c_2$ can be expressed as,
\begin{align}
\label{eq:9}
c_2 = m.c_{2m} + (m-1).(c_{2(m-1)} - c_{2m}) + (m-2).(c_{2(m-2)} - c_{2(m-1)}) + \ldots + 2.(c_{22} - c_{23}) + (c_{21} - c_{22}) \hspace{0.5in}     
\end{align}
If $l$ is the highest power of $2$ such that $k^{'}.2^{l} < n$ ($k^{'} > 0$) and also ($l > m$), then $c_1$ can be written as,
\begin{align}
\label{eq:10}
c_1 = l.c_{1l} + (l-1).(c_{1(l-1)} - c_{1l}) + \ldots + m.(c_{1m} - c_{1(m+1)}) + \ldots + 2.(c_{12} - c_{13}) + (c_{11} - c_{12}) \hspace{1in}     
\end{align}

Because of the inequalities in relations (7) and (8), we see that for all $j = 1, 2, \ldots, m$, the coefficient of $j$ in equation (10) is greater than the coefficeint of $j$ in equation (9). Hence $c_{1} > c_{2}$ and this proves that each of the $5$'s in $5^{c_2}$ in equation (3) are capable of contributing to a trailing zero and the value of $c_2$ is computed following equation (9).

Corollary 1: If $n = 5^{k}$ and $k \geq 1$, then $n!$ has $\frac{5^k - 1}{4}$ trailing zeros.
Proof: Clearly, $m = k$ here, so following equation (9), we have $$c_2 = k.1 + (k-1).( \frac{5^k}{5^{(k-1)}} - 1) + (k-2).(\frac{5^k}{5^{(k-2)}} - \frac{5^k}{5^{(k-1)}}) + \ldots + (k - (k-1)).(\frac{5^k}{5^{(k-(k-1))}} - \frac{5^k}{5^{(k-(k-2))}})$$
\begin{align}
\label{eq:11}
=> c_2 = k + 4.5^0.(k-1) + 4.5^1.(k-2) + 4.5^2.(k-3)+\ldots+4.5^{k-2} \hspace{1.2in}     
\end{align}
\begin{align}
\label{eq:12}
=> \frac{c_2 - k}{4} = 5^0.(k-1) + 5^1.(k-2) + 5^2.(k-3)+\ldots+ 5^{k-2} = S (say) \hspace{0.9in}     
\end{align}
On simplification we get,
\begin{align}
\label{eq:13}
4.S = 5.S - S = (5 + 5^2 + 5^3 + \ldots + 5^{(k-1)}) - (k-1) \hspace{2.5in}     
\end{align}
\begin{align}
\label{eq:14}
\hspace{1.5in} c_{2} - k = \frac{5.(5^{k-1} - 1)}{4} - (k - 1) \hspace{2.8in}     
\end{align}
\begin{align}
\label{eq:15}
\hspace{2.0in} c_{2} = \frac{5^{k} - 1}{4} \hspace{3.8in}     
\end{align}
Corollary 2: If $n = c.5^{k}$, $k \geq 1$ and $c \in {2,3,4}$ then $n!$ has $c.\frac{5^k - 1}{4}$ trailing zeros.
Proof: Since $c \in {2,3,4}$, so the value of $m$ remains same for both $n$ and $\frac{n}{c}$.
Also note that each $c_{2j}$ for $n$ is $c$ times more than that of the $c_{2j}$'s for $\frac{n}{c}$ and the proof follows easily from this observation.

Problems
Prob1: How many zeros does $6250!$ end with?
Solution: We need to first find out highest power $m$ such that $k.5^{m} \leq = 6250$ for some $k > 0$. Note that, $2.5^{5} = 6250$ and hence $m = 5.$ Remember that, $c_{2j} = \lfloor \frac{n}{5^{j}} \rfloor$ for $j = 1, 2, \ldots, 5$. Therefore, $c_{25} = 2$, $c_{24} = 10$, $c_{23} = 50$, $c_{22} = 250$ and $c_{21} = 1250$. Hence, number of trailing zeros in $6250!$ is equal to $$c_{2} = 2.5 + (10 - 2).4 + (50 - 10).3 + (250 - 50).2 + (1250 - 250).1 == 1562$$
Prob2: If $n!$ has $20$ zeros at the end, then what are the possible values of $n$?
Solution: First we need to find out the lowest power $m_{1}$ such that $5^{m_{1}}!$ has more than $20$ trailing zeros, i.e., $c_2$ of $5^{m_{1}}! > 20$. From the corollary1 above, we can easily see that $5^{2}!$ has $6$ trailing zeros and $5^{3}!$ has $31$ trailing zeros.
Usng corollary 2, it can be concluded that $(3.5^{2})! = 75!$ has $18$ trailing zeros and $(4.5^{2})! = 100!$ has $24$ trailing zeros. So, the values of $n$ will be between $75$ and $100$.
Since $75!$ has $18$ trailing zeros, and we are required to find $n$ with $20$ trailing zeros. So, $2$ more $5$ factors we are required to add and this essentially means $75 + 2.5 = 85$ and hence $85!$ will have $20$ trailing zeros. The other possible values of $n$ are $86, 87, 88$ and $89$.