# Number of Divisors/Factors

If one integer can be divided by another integer an exact number of times, then the first number is said to be a*multiple*of the second, and the second number is said to be a

*factor*of the first.

*multiple*of 6 because 6 goes into 48 an exact number of times (8 times in this case). Similarly, 6 is a

*factor*of 48. On the other hand, 48 is not a multiple of 5, because 5 does not go into 48 an exact number of times. So, 5 is not a factor of 48.

Since 20 = 2

^{2}Ã— 5

^{1}

^{}We will take three power of 2 viz., 2

^{0}, 2

^{1 }and 2

^{2}and two powers of 5 viz., 5

^{0}and 5

^{1}.

2

^{0}Ã— 5

^{0}= 1

2

^{0}Ã— 5

^{1}= 5

2

^{1}Ã— 5

^{0}= 2

2

^{1}Ã— 5

^{1}= 10

2

^{2}Ã— 5

^{0}= 4

2

^{2}Ã— 5

^{1}= 20

Now, to find out the number of divisors of any number, we can use the following formula

If

*N*is any number which can be factorised like

*N*=

*a*

^{p}*Ã—*

*b*

^{q}*Ã—*

*c*

^{r}*Ã—â€¦, where*

*a*,

*b*and

*c*are prime numbers.

Number of divisors = (

*p*+ 1) (

*q*+ 1) (

*r*+ 1)â€¦

Example-1

Find the number of divisors of

*N*= 420.Solution

*N*= 420 = 2

^{2}Ã— 3

^{1}Ã— 7

^{1 }Ã— 5

^{1}

So, the number of divisors =

(2 + 1) (1 + 1) (1 + 1) (1 + 1) = 24

Example-2

Find the total number of even and prime divisors of

*N*= 420.Solution

*N*= 420 = 2

^{2}Ã— 3

^{1}Ã— 7

^{1 }Ã— 5

^{1}

Odd divisors will come only if we take zero power of 2 (since, any number multiplied by any power (â‰¥1) of 2 will give us an even number)

So, odd divisors will come if we take

*N*

_{1}= 2

^{0}Ã— 3

^{1}Ã— 7

^{1 }Ã— 5

^{1}

So, number of odd divisors = (0 + 1) (1 + 1) (1 + 1) (1 + 1) = 8

So, total number of even divisors = total number of divisors â€“ number of odd divisors = 24 â€“ 8 = 16

Alternatively, we can also find out the number of even divisors of

*N*= 420 directly (or, in general for any number).

420 = 2

^{2}Ã— 3

^{1}Ã— 7

^{1 }Ã— 5

^{1}

To obtain the factors of 420 which are even, we will not consider 2

^{0}, since 2

^{0}= 1

So, the number of even divisors of 420 = (2) (1 + 1)

(1 + 1) (1 + 1) = 16

(We are not adding 1 in the power of 2, since we are not taking 2

^{0}here, i.e., we are not taking one power of 2.)

Prime divisor = 4 (namely 2, 3, 5 and 7 only)

Example-3

N = 2

^{7}Ã— 3^{5}Ã— 5^{6}Ã— 7^{8}. How many factors of*N*are divisible by 50 but not by 100?Solution

All the factors which are divisible by 50 but not divisible by 100 will have atleast two powers of 5, and one power of 2.

And its format will be 2

So, number of divisors = 1 Ã— 6 Ã— 5 Ã— 9 = 270

Alternatively, this is equal to (Number of factors divisible by 50) â€“ (Number of factors divisible by 100).

And its format will be 2

^{1}Ã— 5^{2+y}.So, number of divisors = 1 Ã— 6 Ã— 5 Ã— 9 = 270

Alternatively, this is equal to (Number of factors divisible by 50) â€“ (Number of factors divisible by 100).

# Condition for two divisors of any number n to be co-prime to each other

Let us see it for*N*= 12

*N*=

*a*

^{p}*Ã—*

*b*

*, will be equal to [(*

^{q}*p*+ 1) (

*q*+ 1) â€“ 1 +

*pq*].

*N*=

*a*

^{p}*Ã—*

*b*

^{q}*Ã—*

*c*

*, then set of co-prime factors can be given by [(*

^{r}*p*+1) (

*q*+ 1) (

*r*+ 1) â€“ 1 +

*pq*+

*qr*+

*pr*+ 3

*pqr*]

Example

Find the set of co-prime factors of the number N = 720.

Solution

720 = 2

^{4}Ã— 3^{2}Ã— 5^{1}Using the formula for three prime factors [(

*p*+ 1) (*q*+ 1) (*r*+ 1) â€“ 1 +*pq*+*qr*+*pr*+ 3*pqr*]^{4}Ã— 3

^{2 }= [(4 + 1) (2 + 1) â€“ 1 + 4.2] = 22

^{22}Ã— 5

^{1 }will give us [(22 + 1) (1 + 1) â€“ 1 + 22.1] = 67

# Sum of Divisors

Like the number of divisors of any number, we can find out the sum of divisors also.If

*N*is any number which can be factorised like

*N*=

*a*

*Ã—*

^{p}*b*

^{q}*Ã—*

*c*

^{r}*Ã—â€¦, where*

*a*,

*b*and

*c*are prime numbers. Then sum of the divisors

=