Formula for finding number of factors :

Let a is positive integer and expressed as product of prime factors.

a = (p_{1})^{x1} . (p_{2})^{x2 }.(p_{3})^{x3} ......... where p_{1}, p_{2}, p_{3}..... are prime numbers and x_{1}, x_{2}, x_{3} are positive integers, then the number of factors of a are (x_{1}+1)(x_{2}+1)(x_{3}+1).....

Example: 8 = 2^{3} means 3 + 1 = 4 factors.

The number of factors of 8 are 4 they are 1, 2, 4, 8.

Q1 : Find the number of factors of the following numbers :

1) 48 2) 72 3) 15, 4) 20 5) 140 6) 120

Solution : 1) 48 = 2 x 2 x 2 x 2 x 3

48 = 2^{4} x 3^{1 }

The number of factors of 48 are (4 + 1) ( 1+ 1) =5 x 2 = 10

Therefore 48 has 10 factors.

2) 72 = 2^{3} x 3^{2 }

The number of factors of 72 are (3 + 1) (2 +1 ) = 4 x 3 = 12

Therefore 72 has 12 factors.

3) 15 = 3^{1} x 5^{1 }

Therefore 15 has (1 + 1)(1 +1 )= 2 x 2 = 4 factors.

4) 20 = 2^{2 }x 5^{1}

Therefore 20 has (2 + 1)(1 + 1) =3 x 2 = 6 factors.

5) 140 = 2^{2 }x 5^{1} x 7^{1 }

Therefore 140 has (2 + 1)( 1 +1)(1 +1) = 3 x 2 x 2 = 12 factors.

6) 120 = 2^{3} x 3^{1} x 5^{1 }

Therefore 120 has (3 + 1)(1 + 1)(1 + 1) = 4 x 2 x 2 = 16 factors.

## Proper and Improper Factors

Proper and Improper factors :

Each whole number has two factors 1 and itself .

Proper factors are factors other than 1 and itself of a number if exist.

Improper factors are factors only 1 and itself of a number .

Note :

i) Every number has improper factors.

ii) Prime numbers has only improper factors.

iii) Composite numbers has both proper and improper factors.

iv) To get number of proper factors subtract 2 from total number of factors.

Q1 : Find the number of proper factors of the following numbers.

a) 56 b) 63 c) 45 d) 94 e) 81 f) 25 h)72

Solution: a) 56 = 2^{3} x 7^{1 }

Total number of factors = (3 + 1)(1 + 1) =4 x 2 = 8

Proper factors are factors other than 1 and itself.

So , number of proper factors = 8 - 2 = 6.

b) 63 = 3^{2}x7^{1 }

Number of proper factors = (2 + 1)(1 +1) - 2

= 3 x 2 - 2

=6 - 2

=4

Therefore 63 has 4 proper factors.

c) 45 = 3^{2} x 5^{1 }

The number of proper factors of 45 are ( 2 + 1)(1 + 1) - 2 = 3 x2 - 2 = 4.

d) 94 = 2^{1} x 47^{1}

The number of proper factors of 94 = (1 + 1)(1 +1) - 2 = 2 x 2 - 2 = 2.

e)81 = 3^{4}

Therefore 81 has (4 +1) -2 = 5 -2 =3 proper factors.

f) 25 = 5^{2}

25 has (2 +1 ) - 2 = 3-2 = 1 proper factor.

g) 72 = 2^{3} x 3^{2}

72 has (3 +1)(2 +1 ) - 2 =4 x 3 -2 =10 proper factors.