 # Exponent of Prime in Factorial

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 Greatest Integer Function with an example. This is important while solving questions from the topic Permutations and Combinations.

Consider n!. It will have prime factors 2, 3, 5, 7..so on.

Let $n!=2^{e_{1}}.3^{e_{2}}.5^{e_{3}}.7^{e_{4}} …$

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 prime in factorial using normal calculation methods. So, we use the following formula to find the exponent of prime in factorial.

## Formula For Exponent of Prime in Factorial

Let n be any positive integer and p be a prime number such that

$n!=p_{1}^{e_{1}}.p_{2}^{e_{2}}.p_{3}^{e_{3}}.p_{4}^{e_{4}} ..$

Exponent of p1 in n! is given by

e1 = [n/p1] + [n/p12] + [n/p13] + …

Here [ . ] denotes the greatest integer function.

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

$8! = 2^{e_{1}}.3^{e_{2}}.5^{e_{3}}…$

Note that [.] denotes 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

### 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 these

Solution:

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.

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

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:

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:

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:

Exponent of 7 in 343! is

1) 57

2) 58

3) 56

4) 55

Solution:

Given n = 343, p1 = 7

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.

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