Question

How many zeroes are there in $100!$ $(100{\text{ factorial)}}$?

Answer 1

Hint:
We know that $100! = (100)(99)(98)............(3)(2)(1)$ and we know that when we multiply the multiple of $5{\text{ and 2}}$ we get the zero at the end of the number. Hence we just need to find the number of pairs of the multiples of $5{\text{ and 2}}$ and get the number of zeros at the end of $100!$

Complete step by step solution:
Here we need to find the number of zeros at the end of the $100!$
For this we need to know what actually factorial means. When we are given $n!$ then it actually means that we need to multiply it with the number preceding it until we get $1$
So we can say that $n! = n(n - 1)(n - 2).........3.2.1$
Here we can put $n = 100$ so we will get:
$100! = (100)(99)(98).........(3)(2)(1)$
We know that when we multiply $2{\text{ and 5}}$ or their multiples we will get at the end or we can say at the unit’s place as zero.
As we are told to find the number of zeros at the end of $100!$
So we need to find the number of multiples of $2{\text{ and 5}}$ which are there between $1{\text{ and 100}}$ and then find how many common pairs of them can be found. So let us firstly find the multiples of $5$
We know that multiples of five between $1{\text{ and 100}}$ are:
$5,10,15,20,............95,100$
Hence we get that there are $20$ multiples of $5$ between $1{\text{ and 100}}$
Now we need to know that in $25,50,75,100$ we have two $5{\text{'s}}$ so we need to count them again. These four $5{\text{'s}}$ are to be counted again. Hence we get number of five’s as:
$20 + 4 = 24$
Now let us see the multiples of $2$ we will see that between $1{\text{ and 100}}$ the multiples of $2$ are:
$2,4,6,8,..........96,98,100$
So we will get $50$ multiples of $2$ between $1{\text{ and 100}}$
Now we know that there are also the $25$ multiples of $4'{\text{s}}$ which give us $25$ more ${\text{2's}}$ and also we have $12$ multiples of $8$ giving $12$ more ${\text{2's}}$
But now we need to see the pairs that can be formed of $2{\text{ and 5}}$
So we will get that there can only be $24$ pairs possible that can give us $0$ at the end.

Hence there will be $24$ zeroes at the end of $100!$

Note:
Here we can also do this type of question with the help of a shortcut where we need to find the number of zeros at the end of $n!$ we can calculate multiples of $5$ by:
Using $\left[ {\dfrac{n}{5}} \right] + \left[ {\dfrac{n}{{{5^2}}}} \right] + \left[ {\dfrac{n}{{{5^3}}}} \right] + ...........$ till we do not get the value as zero and in this we have used the greatest integer function of the number.
Now we can similarly find the multiple of $2$ by:
Using $\left[ {\dfrac{n}{2}} \right] + \left[ {\dfrac{n}{{{2^2}}}} \right] + \left[ {\dfrac{n}{{{2^3}}}} \right] + ...........$ till we do not get the value as zero and in this we have used the greatest integer function of the number.
Now we can easily find the pairs by choosing the smaller value.