Maths@Wordpandit – Basics of Factors

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BASICS OF FACTORS

Test links for this topic at the bottom of the page; scroll to the bottom of the page to take the

tests.

What are factors? What does the term mean? If we go by t he Latin translation of the word, it means

‘who/which acts’. Well, in our mathematical sense, factors are the ones that act for sure, but in this case

they act on numbers.

Mathematically speaking: If any integer, say P, is divisible by another integer say Q an exact number of

times then P is said to be a multiple of Q and Q is the factor of P.

Number of Factors of a given number:

Number of factors can be expressed by following steps:

1. First write down the number in prime factorisation form i.e. in ap b

q c

r(where a ,b,c, are prime numbers

and the p,q,r, are natural numbers as their respective powers.)

2. Number of factors now can be expressed as ( p+1)(q+1)(r+1).

3. Number of factor include 1, and the number itself.

The points above explained through an example:

Let us take an example of a number N = 38491200=26 3

7 5

2

Now observe some facts about the number of factors (we will solve the problem step by step):

Step 1: Prime factorisation, so N=38491200=26 3

7 5

2

Power of 2 as 20

, 21 ,2

2,2

3 ,2

4,2

5,2

6( 6+1=7)ways ,

Power of 3 as 30

, 31 ,3

2,3

3 ,3

4,,3

5,3

6,3

7( 7+1=8)

Power of 5 as 50

, 51 ,5

2( 2+1=3)ways

Step 2: Hence, the number of factors is given by (6+1)(7+1)(2+1)=7x8x3=168

Example 1: Find the number of factors of 24

315

27

2.

Solution: As we can see the above number has 2,3,5,7 which all are prime numbers and they have

4,1,2,2 as their power so the number of factors of the given number are (4+1)(1+1)(2+1)(2+1)= 90

Types of Factors illustrated with Examples

Even factors: Even factors are the factors of number which are divisible by 2

Example 2: Find the number of even factors of 24

315

27

2.

Solution: In this case we have to find number of even factors, an even factor is divisible by 2 or smallest

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power of 2 as 1 not 0, as in the case of total number of factors. Hence a factor must be 2(1 or 2 or 3 or 4)

3(0 or

1 )5

(0 or 1 or 2)7

(0 or 1 or 2)

Hence total number of factors = (4)(1 + 1 )(2 + 1)(2 + 1) = 72

Hence number of even factors of a number N = 2Pa

Qb

rc

sis = p(q + 1)(r + 1)(s + 1)

Odd factors: Odd factors are those factors which are not divisible by 2

Example 3: Find the number of odd factors of24

315

27

2.

Solution: From Example 1 we have seen that total number of factors of N is 240 and from example 2

total number of even factors is 180 hence number of odd factors = 90- 72 = 18

Hence number of odd factors = Total number of factors – Number of even factors

Alternate way: Since odd factors should have power of 2 as 0, the odd factors must be

2(0 )

3(0 or 1 )

5(0 or 1or 3)

7(0 or 1 or 2 )

Total number of odd factors = (1 )(1 + 1)(2 + 1)(2+1) = 18

Hence, number of odd factors of a number N = 2Pa

Qb

rc

s = (q + 1)(r + 1)(s + 1)

Factors that end with zero :


315

27

2 that end with 0 (Or in other words, are divisible by

10)

Solution: If a number is divisible by 10 then it must have minimum power of 2 and 5 as 1.

A factor divisible by 10 must be 2(1 or 2 or 3 or 4)

3(0 or 1 )

5(1 or 2 )

7(0 or 1 or 2 )

Hence, total number of factors divisible by 10 is = (4)(1 + 1)(2)(2 + 1) = 48

Example 5: find the number of factor of 24

315

27

2.that are not divisible by 10?

Solution: We know that the total number of factors are 90 , out of that 48 are divisible by 10 so the

number of factors left which are not divisible by 10 i.e. 90 -48=42

Example 6: find the number of factors of 24

315

27

2, which are divisible by 12?

Since we have to find the number of factors which are divisible by 12, to find this divide the given number

by 12 and then find the number of factors of the quotient.

Divide 24

315

27

2 by 12 ( 2

23

1) = 2

2 5

2 7

2and its number of factors are 3 x 3 x 3=27

Perfect square factors: If a number is perfect square then its prime factors must have even

powers.


315

211

2 that are perfect square?

Solution:We know that for a number to be a perfect square, its factor must have the even number of

powers. All we need to explore is the even powers of the factors in this case.

Therefore the perfect square factors can be2( o or 2 or 4 )

3(0 )

5(0 or 2 )

11(0 or 2 )

Hence, the total number of factors which are perfect square are 3x1x2x2=12

Perfect cube factors: If a number is a perfect cube, then the power of the prime factors should be

divisible by 3.

Example 8: find the number of factors of 293

65

511

8 that are perfect cube?

Solution: If a number is a perfect cube, then the power of the prime factors should be divisible by 3. The

possible options are : 2(0 or 3 or 6 or 9)

3(0 or 3 or 6)

5(0 or 3 )

11(0 or 3 or 6 )

.

Hence, the total number of factors which are perfect cube 4x3x2x3=72.

Example 9: How many factors of 293

65

511

8 are both perfect square and perfect cube?

Solution: If a number is both perfect square and perfect cube then the powers of prime factors

must be divisible by 6. Perfect square factor must be 2(0 or 6)

3(0 or 6)

5(0 )

7(0 or 6)

Hence total number of perfect cube factors are 2 x 2 x 2 = 8

Example 10: How many factors of293

65

511

8 are either perfect squares or perfect cubes but not both?

Solution:

Perfect Square (300) ,Perfect Cube (72), both(8)

Hence required number of factors is 292 + 64 = 356

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Remember

If number of factors is odd then the number is a perfect square

If number of factors is even then number is not a perfect square.

This is because if number is perfect square then p, q, and r are even and hence (p + 1) (q + 1) and

(r+ 1) are odd and so product of these numbers is also an odd number.

Topic Tests: Check your learning of the topic

Factors: Test-1

Factors: Test-2

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Maths@Wordpandit – Basics of Factors

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