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Definition: factors of a number

Factors are the numbers we multiply together to get another number. For example 30=5 x 6, here 5 and 6 are factors of 30. When a number divides another number exactly , then the divisor is called a factor of the dividend, and dividend is called a multiple of the divisor. A factor of a number is an exact divisor of that number.

Factors are usually positive or negative whole numbers (No fractions).

How to find factors of a number?

(i) Factors by Multiplication: To find the factors using the multiplication, write the given number as the product of two numbers in different possible ways. All the numbers that we are using in these products are the factors of the given number.

Example 1. Find the all positive factors of 24.

Since 1 x 24=24 , 2 x 12=24, 3 x 8 =24, 6 x 4=24

So 1, 24, 2, 12, 3, 8, 6 and 4 all are factors of 24.

Note : Also –1, -24, -2, – 12, – 3 , – 8, -6 and -4 are also factors of 24 .

(ii) Factors by Division: To find the factors of a number using division, take all the numbers less than or equal to the given number and divide the given number exactly .

Example 2. Find all positive factors of 96.

We can easily check that each of the number 1, 2, 3 , 4, 6, 8, 16, 24, 48 and 96 completely divides 96. So they are factors of 96.

96 ÷ 1= 96, 96 ÷2 = 48, 96 ÷ 3= 32, 96 ÷ 4= 24, 96 ÷ 6= 16,

96 ÷ 8= 12, 96 ÷ 16= 6, 96 ÷ 24= 4, 96 ÷ 48= 2, 96 ÷ 96= 1

Factors of 96 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48 and 96.

Properties of factors

(i) 1 is factor of every number as 6=6 x 1, 19=19 x 1 , 27=27 x 1 etc.

(ii) Every non-zero number is a factor of itself.

(iii) Every factor of a number is an exact divisor of that number.

(iv) A factor of a non-zero number is less than or equal to the given number.

(v) Number of factors of a given number are finite.

(vi) Every non-zero number is a factor of 0.

(vii) 1 is the only number having one factor , namely itself.

(viii) Every non- zero number other than 1 has at least two factors, namely 1 and itself.

Prime factorisation of a number

If a number is expressed in the form of product of prime numbers , then this form is called the prime factorisation of the given number. e.g. Prime factorisation of 90= 2 x 3 x 3 x 5.

Prime factorisation by building factor tree

We may resolve a number into its prime factors by building factor trees as under:

(i) Find the least prime number by which the given number is divisible. Resolve the number into two factors taking this prime number as one of them .

(ii) Resolve the second factor further into two factors out of which at least one factor is prime .

(iii) Go on splitting the factors till you get all the prime factors.

(iv) Circle all the prime factors.

Example 3. Find the prime factorisation of 84.

Thus, we write prime factorisation of 84 as: 84= 2 x 2 x 3 x 7

Factors Formulas

Let N be the number whose prime factorisation is N= $p_{1}^{_{n_{1}}}\times p_{2}^{_{n_{2}}}\times p_{3}^{_{n_{3}}}\times.............\times p_{n}^{_{n_{n}}}$

, where $p_{1}, p_{2},p_{3}, ............., p_{n}$ are primes and $n_{1}, n_{2}, n_{3}, ........., n_{n}$

are positive integers , then

Total number of factors of N : The total number of positive factors of N are given by $\mathbf{{\color{Red} (n_{1}+1)(n_{2}+1)(n_{3}+1)..........(n_{n}+1)}}$

This total number of factors of N includes 1 and number N itself.

Example 4. Find the total number of positive factors of 540.

Sol . Prime factorisation of 540= $2^{2}\times 3^{3}\times 5$

Total number of positive factors of 540 = (2+1)(3+1)(1+1)= 3 x 4 x 2 = 24

So there are total 24 positive factors of 540.

Example 5. Find the total number of positive factors of 10500 except 1 and itself.

Sol. Prime factorisation of 10500= $2^{2}\times 3^{1}\times 5^{3}\times 7$

Total number of positive factors of 10500= (2+1)(1+1)(3+1)(1+1)= 3 x 2 x 4 x 2 = 48

But we have to exclude 1 and 10500, So there are only 48-2 = 46 positive factors of 10500 except 1 and itself.

Sum of all factors of N: Sum of all factors of N is $\mathbf{{\color{Red} \frac{(p_{1}^{n_{1}+1 }-1)}{(p_{1}-1)}\times \frac{(p_{2}^{n_{2}+1 }-1)}{(p_{2}-1)}\times \frac{(p_{3}^{n_{3}+1 }-1)}{(p_{3}-1)}\times ...........\times \frac{(p_{n}^{n_{n}+1 }-1)}{(p_{n}-1)}}}$

Example 6. Find the sum of factors of 90.

Sol. Prime factorisation of 90= 2 x 3 x 3 x 5 = $2^{1}\times 3^{2}\times 5^{1}$

Here $p_{1}=2, \, p_{2}=3 \, \, p_{3}=5$

and $n_{1}=1, \, n_{2}=2, \, n_{3}=1$

Therefore sum of factors of 90= $(\frac{2^{1+1}-1}{2-1})\times (\frac{3^{2+1}-1}{3-1})\times (\frac{5^{1+1}-1}{5-1})$

= $\frac{3}{1}\times \frac{26}{2}\times \frac{24}{4}=3\times 13\times 6$ = 234

Example 7. The sum of the factors of 1520 except the unity is (a) 3570 (b) 3270 (c) 2730 (d) 3719

Sol. Prime factorisation of 1520= $2^{4}\times 5\times 19$

$\therefore$ Sum of all factors of 1520 = $(\frac{2^{4+1}-1}{2-1})\times (\frac{5^{1+1}-1}{5-1})\times (\frac{19^{1+1}-1}{19-1})$

= 31 x $\frac{24}{4}\times \frac{360}{18}$ = 3720

Since unity is to be excluded, The net sum of the factors= 3720-1 =3719

Product of factors: Product of factors of N is ${\color{Red} \mathbf{N^{(\frac{n}{2})}}}$ where n is the total number of factors of N.

Example 8. Find the product of all the factors of 360.

Sol. Prime factorisation of 360 = $2^{3}\times 3^{2}\times 5^{1}$

$\therefore$ total number of factors =(3+1)(2+1) (1+1)=4 x 3 x 2=24

Thus the product of the factors of 360= $(360)^{\frac{24}{2}}= (360)^{12}$

Example 9. Find the product of 7056.

Sol. Prime factorisation of 7056= $2^{4}\times 3^{2}\times 7^{2}$

$\therefore$

total number of factors of 7056= (4+1)(2+1)(2+1)= 5 x 3 x 3= 45

Thus the product of the factors of 7056= $(7056)^{\frac{45}{2}}$

= $((7056)^{\frac{1}{2}})^{45}$

= $(84)^{45}$ .

Total number of odd and even factors

Let N is a composite number whose prime factors are N= $2^{p}\times b^{q}\times c^{r}\times .....$

Then (i) total number of odd factors = (q+1) x (r+1)…………….

(ii) total number of even factors= p x (q+1) x (r+1)…………….

Example 10: Find the total number of odd and even factors of 180.

Sol. Prime factorisation of 180= $2^{2}\times 3^{2}\times 5$

Number of even factors= p x (q+1) x (r+1), here p=2, q=2, r=1

=2 x (2+1) x (1+1) =2 x 3 x 2 = 12

Number of odd factors = (q+1) x (r+1)

= (2+1) x (1+1) = 3 x 2 =6

Sum of all odd and even factors

Let N is a composite number whose prime factors are N= $2^{p}\times b^{q}\times c^{r}\times .....$

Then Sum of all odd factors = $\mathbf{{\color{Red} \frac{(b^{q+1 }-1)}{(b-1)}\times \frac{(c^{r+1 }-1)}{(c-1)}\times ..........}}$ (we should not consider the power of even prime).