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Question: The exponent of \[3\] in \[100!\] is A.47 B.48 C.49 D.46...

The exponent of 33 in 100!100! is
A.47
B.48
C.49
D.46

Explanation

Solution

Hint: By the exponent of a number in a factorial of another number, we mean the degree of the number by which the factorial can be divided to give natural numbers.

Complete step by step Answer:
Here, we have to find the maximum degree of 33that can divide100100. We will also use the Greatest Integer function to find the exponent.

We know that n!=n×(n1)×(n2)×(n3)....×3×2×1n! = n \times (n - 1) \times (n - 2) \times (n - 3).... \times 3 \times 2 \times 1

So 100!100!can be written as
100!=100×99×98×97....×3×2×1100! = 100 \times 99 \times 98 \times 97.... \times 3 \times 2 \times 1

We know that exponent of a prime number ppin n!n!is
[np]+[np2]+[np3]+[np4]\left[ {\dfrac{n}{p}} \right] + \left[ {\dfrac{n}{{{p^2}}}} \right] + \left[ {\dfrac{n}{{{p^3}}}} \right] + \left[ {\dfrac{n}{{{p^4}}}} \right]….. and so on.

Where [x][x]is the greatest integer less than or equal to xx.

Therefore, the exponent of 33is

[10031]+[10032]+[10033]+[10034] =[1003]+[1009]+[10027]+[10081] =[33.33]+[11.11]+[3.703]+[1.234] =33+11+3+1 =48  \left[ {\dfrac{{100}}{{{3^1}}}} \right] + \left[ {\dfrac{{100}}{{{3^2}}}} \right] + \left[ {\dfrac{{100}}{{{3^3}}}} \right] + \left[ {\dfrac{{100}}{{{3^4}}}} \right] \\\ = \left[ {\dfrac{{100}}{3}} \right] + \left[ {\dfrac{{100}}{9}} \right] + \left[ {\dfrac{{100}}{{27}}} \right] + \left[ {\dfrac{{100}}{{81}}} \right] \\\ = [33.33] + [11.11] + [3.703] + [1.234] \\\ = 33 + 11 + 3 + 1 \\\ = 48 \\\

We only consider the degree to be 44because beyond that the value of the exponent reaches 00, for example: [10035]=[100243]=[0.411]=0\left[ {\dfrac{{100}}{{{3^5}}}} \right] = \left[ {\dfrac{{100}}{{243}}} \right] = [0.411] = 0. Hence, the maximum degree is 44.

The greatest integer function rounds down the real number to its nearest integer less than or equal to the real number.

Hence, the exponent of 33in 100!100!is 4848.

Thus, the answer is option B.

Note: We need to remember that an exponent can be calculated for only prime numbers and we have to use the greatest integer function to solve the problem. If we are given a non-prime number, we will have to prime factorize the number to find the prime numbers and then find the exponent.
Also, we can look at the question in another way.
We can find the number of multiples of different degrees of the prime number in between 1&n1\& n and add all of them to find the exponent.

In this question, there are 3333multiples of 3 in between 1&1001\& 100, 11 multiples of 9, 3 multiples of 27 and 1 multiple of 81. Hence, 33+11+3+1=4833 + 11 + 3 + 1 = 48