# How To Find Sum of Factors of Composite Numbers

Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In the post of * Factors and Multiples of 1 to 100*, you get that the sum of factors of prime numbers and composite numbers are given in the table. In this article, you will learn about

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**FACTORS FORMULA***?.*

**How To Find Sum of Factors of Composite Numbers**Since the factors of a prime number is one and itself, the sum of factors of a prime number is one more than the prime number.

For example: let a prime number is 59. Then, its factors are 1 and 59 only. Thus, the sum of the factors of 59 is 59 + 1 = 60.

It is very simple to find sum of factors of a prime number.

Now, take another example of a composite number 150.

All factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150.

Sum of all factors of 150 is (1 + 2 + 3 + 5 + 6 + 10 + 15 + 25 + 30 + 50 + 75 + 150) = 372.

**Read More About Factors Formulas**

Formula for Total Factors | Formula for Even Factors |

Formula for Odd Factors | Formula for Sum of All Factors |

Formula for Product of All Factors | Factors & Multiples |

Since these numbers are smaller numbers, you get the sum of factors easily by adding all the factors of that number.

But if a bigger composite number, let 4200 is given, then it is very difficult to find the sum of all factors as it is very time consuming process to list all the factors of a bigger composite numbers.

Therefore, you need a short trick to find the sum of all factors of a composite number.

Let N is the composite number whose the sum of all its factors is to be calculated.

Then, its prime factorization is N = α^{a} × β^{b} × γ^{c} × δ^{d }× …, where α, β, γ, δ, … are prime numbers and a, b, c, d, … positive integers.

The factors of α^{a} are 1, α^{1}, α^{2}, α^{3}, … α^{a}. Sum of factors of α^{a} is (1 + α^{1 }+ α^{2 }+ α^{3 }+ … + α^{a}).

The factors of β^{b} are 1, β^{1}, β^{2}, β^{3}, … β^{b}. Sum of factors of β^{b} are (1 + β^{1 }+ β^{2 }+ β^{3 }+ … + β^{b}).

The factors of γ^{c} are 1, γ^{1}, γ^{2}, γ^{3}, … γ^{b}. Sum of factors of γ^{c} are (1 + γ^{1 }+ γ^{2 }+ γ^{3 }+ … + γ^{c}).

and so on…

## Short Trick To Find Sum of All Factors of Composite Numbers

Thus, sum of all factors of N is

S = (1 + α^{1 }+ α^{2 }+ α^{3 }+ … + α^{a}) × (1 + β^{1 }+ β^{2 }+ β^{3 }+ … + β^{b}) × (1 + γ^{1 }+ γ^{2 }+ γ^{3 }+ … + γ^{c}) × …

Using the sum formula of geometric progression, you get short trick to find sum of all factors of a composite numbers as below.

^{a}× β

^{b}× γ

^{c}× … is

[(α

^{a+1}– 1)/(α – 1)] × [(β

^{b+1}– 1)/(β – 1)] × [(γ

^{c+1}– 1)/(γ – 1)] ×…

This sum of all factors of N includes 1 and the number N itself.

## Solved Examples : Sum of All Factors of Composite Numbers

**Ques 1** : Find the sum of all factors of 36.

**Solution** : Prime Factorization of 36 is 36 = 2^{2} × 3^{2}.

Thus, the sum of all factors of 36 is [(2^{2+1} – 1)/(2 – 1)][(3^{2+1} – 1)/(3 – 1)] = 91.

**Ques 2** : Find the sum of all factors of 216.

**Solution** : Prime Factorization of 216 is 216 = 2^{3} × 3^{3}.

Thus, the sum of all factors of 216 is [(2^{3+1} – 1)/(2 – 1)][(3^{3+1} – 1)/(3 – 1)] = 600.

**Ques 3** : Find the sum of all factors of 196.

**Solution** : Prime Factorization of 196 is 196 = 2^{2} × 7^{2}.

Thus, the sum of all factors of 196 is [(2^{2+1} – 1)/(2 – 1)][(7^{2+1} – 1)/(7 – 1)] = 399.

**Ques 4** : Find the sum of all factors of 60.

**Solution** : Prime Factorization of 60 is 60 = 2^{2} × 3 × 5.

Thus, the sum of all factors of 60 is [(2^{2+1} – 1)/(2 – 1)][(3^{1+1} – 1)/(3 – 1)][(5^{1+1} – 1)/(5 – 1)] = 168.

**Ques 5** : Find the sum of all factors of 96.

**Solution** : Prime Factorization of 96 is 96 = 2^{5} × 3.

Thus, the sum of all factors of 96 is [(2^{5+1} – 1)/(2 – 1)][(3^{1+1} – 1)/(3 – 1)] = 252.

**Ques 6** : Find the sum of all factors of 2800.

**Solution** : Prime Factorization of 2800 is 2800 = 2^{4} × 5^{2} × 7.

Thus, the sum of all factors of 2800 is [(2^{4+1} – 1)/(2 – 1)][(5^{2+1} – 1)/(5 – 1)][(7^{1+1} – 1)/(7 – 1)] = 7688.

**Ques 7** : Find the sum of all factors of 16.

**Solution** : Prime Factorization of 16 is 16 = 2^{4}.

Thus, the sum of all factors of 16 is [(2^{4+1} – 1)/(2 – 1)] = 31.

**Ques 8** : Find the sum of all factors of 324.

**Solution** : Prime Factorization of 324 is 324 = 2^{2}× 3^{4}

Thus, the sum of all factors of 324 is [(2^{2+1} – 1)/(2 – 1)][(3^{4+1} – 1)/(3 – 1)] = 847.

**Ques 9** : Find the sum of all factors of 240.

**Solution** : Prime Factorization of 240 is 240 = 2^{4 }× 3^{ }× 5

Thus, the sum of all factors of 240 is [(2^{4+1} – 1)/(2 – 1)][(3^{1+1} – 1)/(3 – 1)][(5^{1+1} – 1)/(5 – 1)] = 744.

**Ques 10** : Find the sum of all factors of 91476.

**Solution** : Prime Factorization of 91476 is 91476 = 2^{2} x 3^{3} x 7^{1} x 11^{2}

Thus, the sum of all factors of 91476 is [(2^{2+1} – 1)/(2 – 1)][(3^{3+1} – 1)/(3 – 1)][(7^{1+1} – 1)/(7 – 1)][(11^{1+1} – 1)/(11 – 1)] = 297,920.

**Ques 11** : Find the sum of all factors of 496.

**Solution** : Prime Factorization of 496 is 496 = 2^{4 }× 31

Thus, the sum of all factors of 496 is [(2^{4+1} – 1)/(2 – 1)][(31^{1+1} – 1)/(31 – 1)] = 992.

**Ques 12** : Find the sum of all factors of 56.

**Solution** : Prime Factorization of 56 is 56 = 2^{3 }× 7

Thus, the sum of all factors of 56 is [(2^{3+1} – 1)/(2 – 1)][(7^{1+1} – 1)/(7 – 1)] = 120.

**Ques 13** : Find the sum of all factors of 19600.

**Solution** : Prime Factorization of 19600 is 19600 = 2^{4} x 5^{2} x 7^{2}

Thus, the sum of all factors of 19600 is [(2^{4+1} – 1)/(2 – 1)][(5^{2+1} – 1)/(5 – 1)][(7^{2+1} – 1)/(7 – 1)] = 54777.

**Ques 14** : Find the sum of all factors of 100.

**Solution** : Prime Factorization of 100 is 100 = 2^{2} x 5^{2}

Thus, the sum of all factors of 100 is [(2^{2+1} – 1)/(2 – 1)][(5^{2+1} – 1)/(5 – 1)] = 217.

**Ques 15** : Find the sum of all factors of 48.

**Solution** : Prime Factorization of 48 is 48 = 2^{4} x 3

Thus, the sum of all factors of 48 is [(2^{4+1} – 1)/(2 – 1)][(3^{1+1} – 1)/(3 – 1)] = 124.

**Ques 16** : Find the sum of all factors of 360.

**Solution** : Prime Factorization of 360 is 360 = 2^{3} x 3^{2} x 5^{1}.

Thus, the sum of all factors of 360 is [(2^{3+1} – 1)/(2 – 1)][(3^{2+1} – 1)/(3 – 1)][(5^{1+1} – 1)/(5 – 1)] = 1170.

**Ques 17** : Find the sum of all factors of 25.

**Solution** : Prime Factorization of 25 is 25 = 5^{2}.

Thus, the sum of all factors of 25 is [(5^{2+1} – 1)/(5 – 1)] = 31.

**Ques 18** : Find the sum of all factors of 2450.

**Solution** : Prime Factorization of 2450 is 2450 = 2^{1} x 5^{2} x 7^{2}.

Thus, the sum of all factors of 2450 is [(2^{1+1} – 1)/(2 – 1)][(5^{2+1} – 1)/(5 – 1)][(7^{2+1} – 1)/(7 – 1)] = 5301.

**Ques 18** : Find the sum of all factors of 720.

**Solution** : Prime Factorization of 720 is 720 = 2^{4} x 3^{2} x 5^{1}.

Thus, the sum of all factors of 720 is [(2^{4+1} – 1)/(2 – 1)][(3^{2+1} – 1)/(3 – 1)][(5^{1+1} – 1)/(5 – 1)] = 2418.

**Ques 18** : Find the sum of all factors of 243.

**Solution** : Prime Factorization of 243 is 243 = 3^{5}.

Thus, the sum of all factors of 243 is [(3^{5+1} – 1)/(3 – 1)] = 364.

**Ques 19** : Find the sum of all factors of 120.

**Solution** : Prime Factorization of 120 is 120 = 2^{3} x 3^{1} x 5^{1}.

Thus, the sum of all factors of 120 is [(2^{3+1} – 1)/(2 – 1)][(3^{1+1} – 1)/(3 – 1)][(5^{1+1} – 1)/(5 – 1)] = 360.

**Ques 20** : Find the sum of all factors of 576.

**Solution** : Prime Factorization of 576 is 576 = 2^{6} x 3^{2}.

Thus, the sum of all factors of 576 is [(2^{6+1} – 1)/(2 – 1)][(3^{2+1} – 1)/(3 – 1)] = 1651.

Formula for Total Factors | Formula for Even Factors |

Formula for Odd Factors | Formula for Sum of All Factors |

Formula for Product of All Factors | Factors & Multiples |