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

Contents

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

- 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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## 1 thought on “Shortcut to Find Number of Factors of a Number | Sum of Factors of a Number”

## Nelson Abalos

(October 1, 2018 - 1:48 pm)Please can you provide me the derivation on the equation of the getting the sum of all factors of “N” and product of all factors of “N”? And can you also provide explanation. Please reply, this is for my study. They wont promote me if i am not able to present this one clearly.