Question

How many zeroes are there at the end of $100!$?

Answer 1

Hint: We first find the process to create zeroes in the multiplication of $100!$. Then we find the prime factorisation of 10 which helps in bringing zeroes. We follow the theorem and find the maximum number of possible zeros.

Complete step by step answer:
We have to find the number of zeros at the end of $100!$.
Zeroes are created by the inclusion of number 10.
So, we have to find how many 10 can we get from the $100!$.
Now, $n!$ denotes the multiplied form of first n natural numbers.
So, $n!=1\times 2\times 3\times 4\times .....\times \left( n-1 \right)\times n$.
Putting $n=100$, we get $100!=1\times 2\times 3\times 4\times .....\times 99\times 100$.
Now we have to find the prime factorisation of 10.
We know $10=2\times 5$.
It means when we are finding the number of 10s, we have in $100!$, we are actually finding the number of 2s and 5s we have in $100!$.
The theorem to find a particular number’s presence in $n!$ is always equal to
$\left[ \dfrac{n}{a} \right]+\left[ \dfrac{n}{{{a}^{2}}} \right]+\left[ \dfrac{n}{{{a}^{3}}} \right]+...+\left[ \dfrac{n}{{{a}^{r}}} \right]$. Here the particular number is a.
Also, the condition for the theorem is that we go to the maximum possible value of r such that ${{a}^{r}} < n$. The function $\left[ {} \right]$ is the Box function where we take the greatest integer less than the number itself.
Now, number of 2s will be
$\begin{align}
  & \left[ \dfrac{100}{2} \right]+\left[ \dfrac{100}{{{2}^{2}}} \right]+\left[ \dfrac{100}{{{2}^{3}}} \right]+\left[ \dfrac{100}{{{2}^{4}}} \right]+\left[ \dfrac{100}{{{2}^{5}}} \right]+\left[ \dfrac{100}{{{2}^{6}}} \right] \\
 & =50+25+12+6+3+1 \\
 & =97 \\
\end{align}$
Now, number of 5s will be
$\begin{align}
  & \left[ \dfrac{100}{5} \right]+\left[ \dfrac{100}{{{5}^{2}}} \right] \\
 & =20+4 \\
 & =24 \\
\end{align}$
So, the maximum number of 10s that is possible to make is 24.
Therefore, the number of zeros in $100!$ is 24.

Note:
We need to remember that the extra number of 2s will never be able to create 10s. The 5s are essential in creating the number 10 as 5 is the prime factorisation of 10. The theorem is only applicable for the number $n!$.