## How to Find Total Number of Factors for Big Numbers Easily | Number System

In *number system* the concept of **factors of numbers** is one of the important sub-topic. In this article, we will discussed about definition of *factors of number,* formulas for finding *number of factors, sum of factors, product of factors, even number of factors, odd number of factors, perfect square factors and perfect cube factors* for any number.

### Definition of Factors of a number:

Factors of a number any number “ P” refers to all the numbers which are exactly divisible on “P” i.e remainder comes to zero. These *factors of number*s are also called **divisors of a number.**** **

**For example:
**

**Factors of the number 9**

- 1 × 9 = 9,
- Also 3 × 3 = 9

So **1, 3, and 9** are factors of 9.

And also -1, -3, and -9 because you get a positive number when you multiply two negatives,

such as (-3)×(-3) = 9

No answer is 1, 3, 9, -1, -3, & -9

**But usually factors of numbers consider only positive numbers**

Note: fractions of numbers also not consider as a factors

#### Formulas related to factors of numbers:

- Total number of factors
- Sum of factors
- Product of factors

Take any number “N” and it is to be covert into **product of prime numbers** (*Prime factorization*) i.e

N = A^{p} x B^{q} x C^{r} here A, B , C are prime numbers and p,q,and r were respective powers of that prime numbers.

**Total numbers of factors for ” N “**= (p + 1)(q +1)(r +1)

**Sum of all factors of “N” **

**Product of all factors of “N”** = ( N )^{Total no. of factors/2}

**Example – 1 :** Find the number of factors of **98** and also find the sum and product of all factors

**Solution :** First write the number 98 into prime factorization

98 = 2 x 49 = 2x 7 x 7

98 = 2^{1} x 7^{2 }Here A = 2 , B = 7 , p= 1 , q = 2

Number of factors for the number 98 = (p + 1)(q +1) = 2 x 3 = 6

Sum of all factors of 98 = = 3 x 57 = 171

Product of all factors of number **98** = (98)^{6/2} = (98)^{3} = 941192

**Example – 2 :** Find the number of factors of **588** and also find the sum and product of all factors

**Solution :** First write the number **588** into prime factorization

588 = 2 x 294 = 2x 2 x 147 = 2 x 2 x 7 x 21 = 2 x 2 x 7 x 7 x 3

588 = 2^{2} x 3^{1} x 7^{2 }Here A = 2 , B = 3 , C = 7 , p= 2 , q = 1 and r =2

Number of factors = (p + 1)(q +1)(r +1) = 3 x 2 x 3 = 18

Sum of all factors of 588 = 7 x 4 x 57 = 1596

Product of all factors of number 588 = (588)^{18/2} = (588)^{9}

#### Another Concepts in Factors of numbers

- How many factors are odd
- How many factors are even
- Number of perfect square factors
- Number of perfect cube factors

**Example – 3** : Find the number of odd, even, perfect square, perfect cube factors of **4500**

**Solution:** First write the number **4500** into prime factorization

4500 = 45 x 100 = 9 x 5 x 10 x 10 = 3 x 3 x 5 x 5 x 2 x 5 x 2

4500 = 2^{2} x 3^{2} x 5^{3} Here consider A = 2 , B = 3 , C = 5 , p= 2 , q = 2 and r = 3

Here identifying that odd number are 3 and 5

Numbers of odd factors of number 4500 = (q + 1 ) (r + 1) = 3 x 4 = 12

Total number of factors = (p + 1)(q +1)(r +1) = 3 x 3 x 4 =36

Numbers of even factors of number = Total number of factors – Numbers of odd factors = 36 – 12 = 24

Number of perfect square factors of number 4500 = 2 x 2 x 2 = 8

( 2^{2} 2^{0} , 2^{2 }; 3^{2} 3^{0} , 3^{2 }& 5^{2} 5^{0} , 5^{2 })

Number of perfect cube factors of number 4500 = 1 x 1 x 2 = 2

( 2^{2} 2^{0}^{ }3^{2} 3^{0} , ^{ }& 5^{2} 5^{0} , 5^{3 })

**Example – 4 :** Find the number of odd, even, perfect square, perfect cube factors of **5040**

**Solution:** First write the number** 5040** into prime factorization

5040 = 504 x 10 = 4 x 126 x 5 x 2 = 2 x 2 x 18 x 7 x 5 x 2 = 2 x 2 x 3 x 3 x 2 x 7 x 5 x 2

5040 = 2^{4} x 3^{2} x 7^{1} Here consider A = 2 , B = 3 , C = 7 , p= 4 , q = 2 and r = 1

Here identifying that odd number are 3 and 7

Numbers of odd factors of number 5040 = (q + 1 ) (r + 1) = 3 x 2 = 6

Total number of factors = (p + 1)(q +1)(r +1) = 5 x 3 x 2 = 30

Numbers of even factors of number = Total number of factors – Numbers of odd factors = 30 – 6 = 24

Number of perfect square factors of number 5040 = 3 x 2 x 1 = 6

( 2^{2} 2^{0} , 2^{2}, 2^{4 }; 3^{2} 3^{0} , 3^{2 }& 7^{1} 7^{0} ^{ })

Number of perfect cube factors of number 5040 = 2 x 1 x 1 = 2

( 2^{2} 2^{0} , 2^{3}^{ }; 3^{2} 3^{0} , ^{ }& 7^{2} 7^{0} ^{ })

**Related Topics :**

Rules for Divisibility of numbers

GCD and LCM Problems & Solutions

Formulas for Sum of n Consecutive numbers

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