The exponent of prime in factorial is an important topic for the JEE exam. In this article, we learn how to find the exponent of prime in factorial by using the formula involving the greatest integer function with an example. This is important while solving questions on the topic of permutations and combinations.
Consider n!. It will have prime factors 2, 3, 5, 7..so on.
Let
Here, we have written factorial notation as the multiples of the prime numbers. Here, e1 is the exponent of prime number 2. e2 is the exponent of prime number 3. e3 is the exponent of prime number 5.
For example, 3! = 21. 31
4! = 1×2×3×4 = 23. 31
5! = 1×2×3×4×5 = 23. 31. 51
In the case of bigger numbers, it is difficult for us to find the exponent of the prime in factorial using normal calculation methods. So, we use the following formula to find the exponent of the prime in factorial.
The Formula for Exponent of Prime in Factorial
Let n be any positive integer and p be a prime number, such that
The exponent of p1 in n! is given by
e1 = [n/p1] + [n/p12] + [n/p13] + …
Here, [ . ] denotes the greatest integer function.
The exponent of p2 in n! is given by
e2 = [n/p2] + [n/p22] + [n/p23] + …
This formula can be used to find the highest power of a prime number in a factorial.
Consider the example 8! = 1×2×3×4×5×6×7×8
Note that [.] denotes a greater integer function.
Here, e1 = [8/21] + [8/22] + [8/23] + [8/24] + ..
= 4 + 2 + 1 + 0
= 7
Now, e2 = [8/31] + [8/32] + [8/33] + …
= 2 + 0
= 2
e3 = [8/51] + [8/52]+…
= 1 + 0
= 1
Also Read
Solved Examples on Exponent of Prime in Factorial
Example 1:
Find the exponent of 2 in 100!
1) 98
2) 99
3) 97
4) None of the above
Solution:
The exponent of p1 in n! is given by
e1 = [n/p1] + [n/p12] + [n/p13] + …
Here, [ . ] denotes the greatest integer function.
We have n = 100 and p1 = 2
e1 = [100/2] + [100/22] + [100/23] + [100/24] + [100/25] + [100/26] + [100/27]+ …
= 50 + 25 + 12 + 6 + 3 + 1 + 0
= 97
Hence, the exponent of 2 in 100! is 97.
Option 3 is the answer.
Example 2:
Find the maximum value of n, such that 3n divides 100!.
1) 46
2) 47
3) 48
4) 49
Solution:
Given n = 100, p1 = 3
The exponent of p1 in n! is given by
e1 = [n/p1] + [n/p12] + [n/p13] + …
Here, [ . ] denotes the greatest integer function.
e1 = [100/3] + [100/32] + [100/33] + [100/34] + [100/35] +..
= 33 + 11 + 3 + 1 + 0
= 48
So the exponent of 3 is 48.
Therefore, 3n = 348
Hence, option 3 is the answer.
Example 3:
The exponent of 2 in 144! is
1) 143
2) 144
3) 121
4) 142
Solution:
The exponent of p1 in n! is given by
e1 = [n/p1] + [n/p12] + [n/p13] + …
Here, [ . ] denotes the greatest integer function.
e1 = [144/2] + [144/22] + [144/23] + [144/24] + [144/25] + [144/26] + [100/27]+ [100/28]+ …
= 72 + 36 + 18 + 9 + 4 + 2 + 1 + 0
= 142
Hence, option 4 is the answer.
Example 4:
What is the highest power of 7 that can divide 5000! without leaving a remainder?
1) 833
2) 835
3) 831
4) 832
Solution:
The exponent of p1 in n! is given by
e1 = [n/p1] + [n/p12] + [n/p13] + …
Here, [ . ] denotes the greatest integer function.
e1 = [5000/7] + [5000/72] + [5000/73] + [5000/74] + [5000/75] + [5000/76] + …
= 714 + 102 + 14 + 2 + 0
= 832
Hence, option 4 is the answer.
Example 5:
The exponent of 7 in 343! is
1) 57
2) 58
3) 56
4) 55
Solution:
Given n = 343, p1 = 7
The exponent of p1 in n! is given by
e1 = [n/p1] + [n/p12] + [n/p13] + …
Here, [ . ] denotes the greatest integer function.
e1 = [343/7] + [343/72] + [343/73] + [343/74] + …
= 49 + 7 + 1 +0
= 57
Hence, option 1 is the answer.
Related videos
Exponent of Prime in Factorial
Greatest Integer Function
Frequently Asked Questions
What do you mean by the factorial of a number?
The factorial of a number is the function that multiplies the number by every natural number below it till 1. For example, 3! = 3×2×1.
How do you find the exponent of prime in a factorial?
Let n be any positive integer and p be a prime number. Exponent of p1 in n! is given by e1 = [n/p1] + [n/p12] + [n/p13] + … [ . ] denotes the greatest integer function.
What is the value of zero factorial?
The value of zero factorial is 1.
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