Question

The number of zeros at the end of \[108!\] is
A.10
B.13
C.25
D.26

Answer 1

Hint: Here, we are required to find the number of zeros which are present at the end of the given factorial. We will use the fact that we get a zero when 5 is multiplied by 2. We will observe which number would require lesser power to give us the required number of trailing zeros. Hence, we will divide by either 2 or 5 accordingly using the formula. We will simplify it further to get the required answer

Complete step-by-step answer:
We have to find the number of zeros at the end of \[108!\]
As we know, if \[n\] is the given natural number, then, its factorial will be written as:
\[n! = n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right) \times ......3 \times 2 \times 1\]
This means that a factorial is the product of all the integers which are less than or equal to the given integer but which are not less than 1.
Now, since we have to find the number of ‘zeros’, then we know that the product of 5 and 2 gives us zero at the end of a number.
Hence, we will use the formula \[\sum\limits_{k \ge 1} {\left\lfloor {\dfrac{n}{{{5^k}}}} \right\rfloor } \].
This means that we will find the number of exponents of 5 in \[108!\] i.e. \[{E_5}\left\lfloor {108!} \right\rfloor \]
Hence, this can be written as:
\[{E_5}\left\lfloor {108!} \right\rfloor = \left\lfloor {\dfrac{{108}}{5}} \right\rfloor + \left\lfloor {\dfrac{{108}}{{{5^2}}}} \right\rfloor + \left\lfloor {\dfrac{{108}}{{{5^3}}}} \right\rfloor \]
\[ \Rightarrow {E_5}\left\lfloor {108!} \right\rfloor = \left\lfloor {\dfrac{{108}}{5}} \right\rfloor + \left\lfloor {\dfrac{{108}}{{25}}} \right\rfloor + \left\lfloor {\dfrac{{108}}{{125}}} \right\rfloor \]
Also, we should keep in mind that we have to find the greatest integer possible.
\[ \Rightarrow {E_5}\left\lfloor {108!} \right\rfloor = 21 + 4 + 0 = 25\]
Hence, the number of zeros at the end of \[108!\] is 25.
Therefore, option C is the correct answer.

Note: To find the exact number of zeros which are present at the end of a factorial, we should know how many times the given number will be divisible by 10. This number of trailing zeros can be found by using the above method, i.e. dividing by exponents of 5 or 2; whichever has a lesser power.
A quick way to find the number of trailing zeros is first of all, divide the given number by 5:
\[\left\lfloor {\dfrac{{108}}{5}} \right\rfloor = 21\]
(Again, taking into consideration, the greatest integer possible)
Now, we can simply divide 21 again by 5,
\[ \Rightarrow \left\lfloor {\dfrac{{21}}{5}} \right\rfloor = 4\]
Now, adding these two numbers, we get, \[21 + 4 = 25\], which is the required answer.