FACTORS

Previously, we studied about factors, total number of factors, even and odd factors. Now, we’ll proceed further with more concepts related to factors.

Perfect Square Factors:

If a number is perfect square then its prime factors must have even powers.

Consider a perfect square: –

You can observe that power of prime factors will get multiplied by 2. Thus, the powers of prime factor in any perfect square will always be even.

Let us understand this with the help of an example.

Example: Find the number of factors of 58800 which  are perfect square?

Solution:

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

We first factorize 58800.

58800 = 2⁴ 3¹5²7²

Hence perfect square factors must have

Power of 2 which will be taken into account = >   2^{(0 or 2 or 4)}—– 3 factors

Power of 3 which will be taken into account = >   3^{( 0 )}     —–   1  factor

Power of 5 which will be taken into account = >   5^{(0 or 2)} ——- 2 factors

Power of 7 which will be taken into account = >   7^{(0 or 2)} — 2 factors

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

Note:

1. If number of factors is odd then the number is a perfect square and vice versa is also true i.e. if a number is a perfect square then number of factors is 

This is because if number is a 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.

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

Perfect Cube Factors:

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

Consider a perfect cube number :-

So we can see that power of prime factors will always be a multiple of 3.

Let us understand this with the help of an example.

Example: Find the number of factors of 2⁹3⁶5⁵11⁸ that are perfect cube.

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

Power of 2 which will be taken into account = >    2^{(0 or 3 or 6or 9)}—– 4 factors

Power of 3 which will be taken into account = >   3^{(0 or 3 or 6)}  —–  3  factors

Power of 5 which will be taken into account = >   5^{(0 or 3}  ——- 2 factors

Power of 11 which will be taken into account = >   11^{(0 or 3 or 6 )}— 3 factors

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

Perfect Square And Perfect Cube:

If a number is both  perfect square and perfect cube then the powers of prime factors must be divisible by 6.

Consider a number which is both perfect square and perfect cube .

So we know the resulting power of that prime number will ultimately be multiple of both 2 and 3.

Let us understand this more with the help of an example.

Example 1: How many factors of 2⁹3⁶5⁵11⁸ 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.Hence both perfect square and perfect cube must have

2^{(0 or 6)}—– 2 factors

3^{(0 or 6)} —– 2 factors

5⁽⁰⁾——- 1 factor

11^{(0 or 6)}— 2 factors

Hence total number of such factors are 2x2x1x2=8

Example 2: How many factors of 2⁹3⁶5⁵11⁸ are either perfect squares or perfect cubes but not both?

Solution:

Let A denote a set of numbers that are perfect squares.

If a number is a perfect square, then the power of the prime factors should be divisible by 2. Hence perfect square factors must have

2^{(0 or 2 or 4 or 6 or 8)}—– 5 factors

3^{(0 or 2 or 4 or 6)}  —– 4 factors

5^{(0 or 2or 4 )}——- 3 factors

11^{(0 or 2or 4 or6 or 8 )}— 5 factors

Hence, the total number of factors which are perfect square i.e. n(A)=5x4x3x5=300

Let B denote a set of numbers that are perfect cubes

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

2^{(0 or 3 or 6or 9)}—– 4 factors

3^{(0 or 3 or 6)}  —–  3  factors

5^{(0 or 3)}——- 2 factors

11^{(0 or 3 or 6 )}— 3 factors

Hence, the total number of factors which are perfect cube i.e. n(B)=4x3x2x3=72

If a number is both perfect square and perfect cube then the powers of prime factors must be divisible by 6.Hence both perfect square and perfect cube must have:

2^{(0 or 6)}—– 2 factors

3^{(0 or 6)} —– 2 factors

5⁽⁰⁾——- 1 factor

11^{(0 or 6)}— 2 factors

Hence total number of such factors are i.e.n(A B)=2x2x1x2=8

We are asked to calculate which are either perfect square or perfect cubes i.e.

n(A U B )= n(A) + n(B) – n(A B)

=300+72 – 8

=364

Hence required number of factors is364.

Prime And Composite Factors:

Natural numbers have 3 kind of factors:

1. Unity ( 1)
2. Prime factors
3. Composite factors

Let us discuss all one by one:

1 is always a factor of every number as it can divide any number.

Prime factors:-  If N is a number then every prime number that can divide it will be its factor.

Composite number: – Composite numbers are made of prime numbers. We can multiply any two or more prime numbers and get composite numbers. So every composite number will have a unique set of prime numbers.

Example:  Find prime factors, composite factors and total factors of 6300.

Solution:

Total factors :- (2+1)(2+1)(2+1)(1+1) =54

Prime factors :-  2,3,5,7

Unity will be a factor too.

Composite factors= total factors – number of prime factors – 1

Composite factors = 54 -4-1 =49

 Let’s try some questions based on the above-discussed concepts.

EXERCISE :-

Question1 : Find the number of factors of 4900  which  are perfect square.

(1) 8

(2) 32

(3) 16

(4) 64

Question 2 : Find the number of factors of 1331000 that are perfect cube.

(1) 64

(2) 32

(3) 16

(4) 8

Question 3 : How many factors of  are both perfect square and perfect cube?

(1) 8

(2) 12

(3) 16

(4) 48

Question 4 : How many factors of  are  perfect square?

(1) 80

(2) 32

(3) 56

(4) 112

Question 5 : How many factors of  are perfect cube?

(1) 9

(2) 32

(3) 27

(4) 81