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Question: The exponent of 12 in \(100!\) is A) 32 B) 48 C) 97 D) None of these...

The exponent of 12 in 100!100! is
A) 32
B) 48
C) 97
D) None of these

Explanation

Solution

Here, we will rewrite the given factorial in the form of the prime powers. We will then use the formula of the exponent of any prime number in factorial to find the exponent of 2 and 3 in the given factorial. We will solve it further to find the exponent in the given factorial.

Formula Used:
The exponent of any prime number pp in n!n! is given by the formula [np]+[np2]+[np3]+.....+[npk]\left[ {\dfrac{n}{p}} \right] + \left[ {\dfrac{n}{{{p^2}}}} \right] + \left[ {\dfrac{n}{{{p^3}}}} \right] + ..... + \left[ {\dfrac{n}{{{p^k}}}} \right] where kk must be chosen such that pk+1>n{p^{k + 1}} > n and [.]\left[. \right] represents the greatest integer function.

Complete step by step solution:
We are given a factorial 100!100!.
We will write the given factorial in the form of prime powers as 12=22×312 = {2^2} \times 3.
First, we will consider the factorial in terms of the power of 2.
Now using the formula [np]+[np2]+[np3]+.....+[npk]\left[ {\dfrac{n}{p}} \right] + \left[ {\dfrac{n}{{{p^2}}}} \right] + \left[ {\dfrac{n}{{{p^3}}}} \right] + ..... + \left[ {\dfrac{n}{{{p^k}}}} \right], we get
The exponent of 2 in 100!=[1002]+[10022]+[10023]+[10024]+[10025]+[10026]100! = \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]
Applying the exponents on the terms of the denominator, we get
\Rightarrow The exponent of 2 in 100!=[1002]+[1004]+[1008]+[10016]+[10032]+[10064]100! = \left[ {\dfrac{{100}}{2}} \right] + \left[ {\dfrac{{100}}{4}} \right] + \left[ {\dfrac{{100}}{8}} \right] + \left[ {\dfrac{{100}}{{16}}} \right] + \left[ {\dfrac{{100}}{{32}}} \right] + \left[ {\dfrac{{100}}{{64}}} \right]
Simplifying the fractions, we get
\Rightarrow The exponent of 2 in 100!=[50]+[25]+[12.5]+[6.25]+[3.125]+[1.5625]100! = \left[ {50} \right] + \left[ {25} \right] + \left[ {12.5} \right] + \left[ {6.25} \right] + \left[ {3.125} \right] + \left[ {1.5625} \right]
Since [.]\left[ . \right] represents the greatest integer function, we get
\Rightarrow The exponent of 2 in 100!=50+25+12+6+3+1100! = 50 + 25 + 12 + 6 + 3 + 1
Adding the terms, we get
\Rightarrow The exponent of 2 in 100!=97100! = 97
Since the exponent of 22 is in the power of 22, we get
\Rightarrow The exponent of 2 in 100!=972100! = \dfrac{{97}}{2}
Dividing the terms, we get
\RightarrowThe exponent of 2 in 100!=48.548100! = 48.5 \approx 48
Now using the formula [np]+[np2]+[np3]+.....+[npk]\left[ {\dfrac{n}{p}} \right] + \left[ {\dfrac{n}{{{p^2}}}} \right] + \left[ {\dfrac{n}{{{p^3}}}} \right] + ..... + \left[ {\dfrac{n}{{{p^k}}}} \right], we get
\Rightarrow The exponent of 3 in 100!=[1003]+[10032]+[10033]+[10034]100! = \left[ {\dfrac{{100}}{3}} \right] + \left[ {\dfrac{{100}}{{{3^2}}}} \right] + \left[ {\dfrac{{100}}{{{3^3}}}} \right] + \left[ {\dfrac{{100}}{{{3^4}}}} \right]
Applying the exponents on the terms of the denominator, we get
\Rightarrow The exponent of 3 in 100!=[1003]+[1009]+[10027]+[10081]100! = \left[ {\dfrac{{100}}{3}} \right] + \left[ {\dfrac{{100}}{9}} \right] + \left[ {\dfrac{{100}}{{27}}} \right] + \left[ {\dfrac{{100}}{{81}}} \right]
Simplifying the fractions, we get
\Rightarrow The exponent of 3 in 100!=[33.33]+[11.11]+[3.7037]+[1.2345]100! = \left[ {33.33} \right] + \left[ {11.11} \right] + \left[ {3.7037} \right] + \left[ {1.2345} \right]
Since [.]\left[ . \right] represents the greatest integer function, we get
\Rightarrow The exponent of 3 in 100!=33+11+3+1100! = 33 + 11 + 3 + 1
Adding the terms, we get
\Rightarrow The exponent of 3 in 100!=48100! = 48
We will get the exponent as 12 if the power of 2 occurs twice and the power of 3 occurs once.
Thus 48 times the exponent of 12 in the factorial 100!100!.
Therefore, the exponent of 12 in 100!100! is 48.
**
Thus, option (B) is the correct answer.**

Note:
We should note that the exponent of any prime number pp in n!n! is applicable only in the case of prime numbers. If the given number is a composite number then it has to be written in the form of a prime number. A prime is a number which has only 2 factors that is the number itself and 1. The function that is rounding off the real number down to the integer less than the number is known as the greatest integer function.