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# Factor Formula | How To Find Product of All Factors of a Number

Hi Students, Welcome to **Amans Maths Blogs (AMB)**. You learnt the concepts as ‘How to Find Number of Factor of Any Number‘, ‘How to Find Number of Even Factors‘, ‘How to Find Number of Odd Factors‘, ‘How to Find Number of Sum of Factors of Any Number‘. In this article, you will learn about * FACTORS FORMULA* |

*In the post of*

**How To Find Product of All Factors of a Number.***, you get that the sum of factors of prime numbers and composite numbers are given in the table.*

**Factors and Multiples of 1 to 100**Let start with an example * how to find the product of factors or divisors of a composite number*.

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 |

You know that the number of factors of a perfect square number is odd and other numbers have even number of factors.

Suppose a number is 120 (which is a non perfect square), you need to find the product of its divisors.

First, you need to find all of its factors. The factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. There are 16 factors of 120.

Now, the required product is

1 × 2 × 3 × 4 × 5 × 6 × 8 × 10 × 12 × 15 × 20 × 24 × 30 × 40 × 60 × 120.

To simplify this product, you need to proceed as below.

1 × 2 × 3 × 4 × 5 × 6 × 8 × 10 × 12 × 15 × 20 × 24 × 30 × 40 × 60 × 120

= (1 × 120) × (2 × 30) × (3 × 40) × (4 × 30) × (5 × 24) × (6 × 20) × (8 × 15) × (10 × 12)

= 120 × 120 × 120 × 120 × 120 × 120 × 120 × 120

= (120)^{8}

= (120)^{16/2}

= (120)^{(Number of Factors of 120)2}

Now, suppose another number is 1225 (which is a perfect square 35^{2} = 1225), you need to find the product of its divisors.

First, you need to find all of its factors. The factors of 1225 are 1, 5, 7, 25, 35, 49, 175, 245, 1225. There are 9 factors of 1225.

Now, the required product is

1 × 5 × 7 × 25 × 35 × 49 × 175 × 245 × 1225.

To simplify this product, you need to proceed as below.

1 × 5 × 7 × 25 × 35 × 49 × 175 × 245 × 1225

= (1 × 1225) × (5 × 245) × (7 × 175) × (25 × 49) × 35

= 1225 × 1225 × 1225 × 1225 × 35

= 1225 × 1225 × 1225 × 1225 × (1225)^{1/2}

= (1225)^{4} × (1225)^{1/2}

= (1225)^{4 + 1/2}

= (1225)^{9/2}

= (1225)^{(Number of Factors of 1225)/2}

On the basis of these two above examples, you can conclude the following.

Suppose N is a number whose the product of its divisors to be find. The prime factorization is α^{a} × β^{b} × γ^{c} × δ^{d }× …., where α, β, γ, δ,… are the prime numbers.

Then,

Total Number of Factors of N = α^{a} × β^{b} × γ^{c} × δ^{d }× … is (a + 1)(b + 1)(c + 1)(d + 1)…

Thus,

^{(Number of Divisors of N)/2}

= N

^{{(a + 1)(b + 1)(c + 1)(d + 1)…} / 2}

This product of factors of N includes 1 and the number N itself.

And,

= α × β × γ × δ

^{ }×

## Solved Examples: Find Product of Factors of Number

**Ques 1** : Find the product of factors of 6.

**Solution** : Prime Factorization of 6 is 6 = 2^{1} × 3^{1}.

Then, Number of Factors of 6 is (1 + 1) × (1 + 1) = 4

Thus, the product of all factors of 6 is 6^{(Number of Factors of 6)/2} = 6^{4/2} = 6^{2} = 36

**Ques 2** : Find the product of factors of 8.

**Solution** : Prime Factorization of 8 is 8 = 2^{3}.

Then, Number of Factors of 8 is (3 + 1) = 4

Thus, the product of all factors of 8 is 8^{(Number of Factors of 8)/2} = 8^{4/2} = 8^{2} = 64

**Ques 3** : Find the product of factors of 10.

**Solution** : Prime Factorization of 10 is 10 = 2^{1} × 5^{1}.

Then, Number of Factors of 10 is (1 + 1) × (1 + 1) = 4

Thus, the product of all factors of 10 is 10^{(Number of Factors of 10)/2} = 10^{4/2} = 10^{2} = 100

**Ques 4** : Find the product of factors of 15.

**Solution** : Prime Factorization of 15 is 15 = 3^{1} × 5^{1}.

Then, Number of Factors of 15 is (1 + 1) × (1 + 1) = 4

Thus, the product of all factors of 15 is 15^{(Number of Factors of 15)/2} = 15^{4/2} = 15^{2} = 225

**Ques 5** : Find the product of factors of 16.

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

Then, Number of Factors of 16 is (4 + 1) = 5

Thus, the product of all factors of 16 is 16^{(Number of Factors of 16)/2} = 16^{5/2} = (4^{2})^{5/2} = 4^{5} = 1024

**Ques 6** : Find the product of factors of 18.

**Solution** : Prime Factorization of 18 is 18 = 2^{1} × 3^{2}.

Then, Number of Factors of 18 is (1 + 1) × (2 + 1) = 6

Thus, the product of all factors of 18 is 18^{(Number of Factors of 18)/2} = 18^{6/2} = 18^{3} = 5832

**Ques 7** : Find the product of factors of 25.

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

Then, Number of Factors of 25 is (2 + 1) = 3

Thus, the product of all factors of 25 is 25^{(Number of Factors of 25)/2} = 25^{3/2} = (5^{2})^{3/2} = 5^{3} = 125

**Ques 8** : Find the product of factors of 30.

**Solution** : Prime Factorization of 30 is 30 = 2^{1} × 3^{1} × 5^{1}

Then, Number of Factors of 30 is (1 + 1) × (1 + 1) × (1 + 1) = 8

Thus, the product of all factors of 30 is 30^{(Number of Factors of 30)/2} = 30^{8/2} = 30^{4} = 810000

**Ques 9** : Find the product of factors of 32.

**Solution** : Prime Factorization of 32 is 32 = 2^{5}.

Then, Number of Factors of 32 is (5 + 1) = 6

Thus, the product of all factors of 32 is 32^{(Number of Factors of 32)/2} = 32^{6/2} = 32^{3} = 32678

**Ques 10** : Find the product of factors of 34.

**Solution** : Prime Factorization of 34 is 34 = 2^{1} × 17^{1}

Then, Number of Factors of 34 is (1 + 1) × (1 + 1) = 4

Thus, the product of all factors of 34 is 34^{(Number of Factors of 34)/2} = 34^{4/2} = 34^{2} = 1156

**Ques 11** : **Find the product of factors of 35.**

**Solution** : Prime Factorization of 35 is 35 = 5^{1} × 7^{1}

Then, Number of Factors of 35 is (1 + 1) × (1 + 1) = 4

Thus, the product of all factors of 35 is 35^{(Number of Factors of 35)/2} = 35^{4/2} = 35^{2} = 1225

**Ques 12** : Find the product of factors of 36.

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

Then, Number of Factors of 36 is (2 + 1) × (2 + 1) = 9

Thus, the product of all factors of 36 is 36^{(Number of Factors of 36)/2} = 36^{9/2} = (6^{2})^{9/2} = 6^{9}

**Ques 13** : Find the product of factors of 39.

**Solution** : Prime Factorization of 39 is 39 = 3^{1} × 13^{1}

Then, Number of Factors of 39 is (1 + 1) × (1 + 1) = 4

Thus, the product of all factors of 39 is 39^{(Number of Factors of 39)/2} = 39^{4/2} = 39^{2} = 1521

**Ques 14** : Find the product of factors of 40.

**Solution** : Prime Factorization of 40 is 40 = 2^{3} × 5^{1}

Then, Number of Factors of 40 is (3 + 1) × (1 + 1) = 8

Thus, the product of all factors of 40 is 40^{(Number of Factors of 40)/2} = 40^{8/2} = 40^{4} = 2560000

**Ques 15** : Find the product of factors of 42.

**Solution** : Prime Factorization of 42 is 42 = 2^{1} × 3^{1 }× 7^{1}

Then, Number of Factors of 42 is (1 + 1) × (1 + 1) × (1 + 1) = 8

Thus, the product of all factors of 42 is 42^{(Number of Factors of 42)/2} = 42^{8/2} = 42^{4}

**Ques 16** : Find the product of factors of 44.

**Solution** : Prime Factorization of 44 is 44 = 2^{2} × 11^{1}

Then, Number of Factors of 44 is (2 + 1) × (1 + 1) = 6

Thus, the product of all factors of 44 is 44^{(Number of Factors of 44)/2} = 44^{6/2} = 44^{3}

**Ques 17** : Find the product of factors of 48.

**Solution** : Prime Factorization of 48 is 48 = 2^{4} × 3^{1}

Then, Number of Factors of 48 is (4 + 1) × (1 + 1) = 10

Thus, the product of all factors of 48 is 48^{(Number of Factors of 48)/2} = 48^{10/2} = 48^{5}

**Ques 18** : Find the product of factors of 50.

**Solution** : Prime Factorization of 50 is 50 = 2^{1} × 5^{2}

Then, Number of Factors of 50 is (1 + 1) × (2 + 1) = 6

Thus, the product of all factors of 50 is 50^{(Number of Factors of 50)/2} = 50^{6/2} = 50^{3 }= 125000

**Ques 19** : Find the product of factors of 54.

**Solution** : Prime Factorization of 54 is 54 = 2^{1} × 3^{3}

Then, Number of Factors of 54 is (1 + 1) × (3 + 1) = 8

Thus, the product of all factors of 54 is 54^{(Number of Factors of 54)/2} = 54^{8/2} = 54^{4 }

**Ques 20** : Find the product of factors of 56.

**Solution** : Prime Factorization of 56 is 56 = 2^{3} × 7^{1}

Then, Number of Factors of 56 is (3 + 1) × (1 + 1) = 8

Thus, the product of all factors of 56 is 56^{(Number of Factors of 56)/2} = 56^{8/2} = 56^{4 }

**Ques 21** : Find the product of factors of 60.

**Solution** : Prime Factorization of 60 is 60 = 2^{2} × 3^{1 }× 5^{1}

Then, Number of Factors of 60 is (2 + 1) × (1 + 1) × (1 + 1) = 12

Thus, the product of all factors of 60 is 60^{(Number of Factors of 60)/2} = 60^{12/2} = 60^{6}

**Ques 22** : Find the product of factors of 62.

**Solution** : Prime Factorization of 62 is 62 = 2^{1} × 31^{1}

Then, Number of Factors of 62 is (1 + 1) × (1 + 1) = 4

Thus, the product of all factors of 62 is 62^{(Number of Factors of 62)/2} = 62^{4/2} = 62^{2}

**Ques 23** : Find the product of factors of 63.

**Solution** : Prime Factorization of 63 is 63 = 3^{2} × 7^{1}

Then, Number of Factors of 63 is (2 + 1) × (1 + 1) = 6

Thus, the product of all factors of 63 is 63^{(Number of Factors of 63)/2} = 63^{6/2} = 63^{3}

**Ques 24** : Find the product of factors of 68.

**Solution** : Prime Factorization of 68 is 63 = 2^{2} × 17^{1}

Then, Number of Factors of 68 is (2 + 1) × (1 + 1) = 6

Thus, the product of all factors of 68 is 68^{(Number of Factors of 68)/2} = 68^{6/2} = 68^{3}

**Ques 25** : Find the product of factors of 69.

**Solution** : Prime Factorization of 69 is 69 = 3^{1} × 23^{1}

Then, Number of Factors of 69 is (1 + 1) × (1 + 1) = 4

Thus, the product of all factors of 69 is 69^{(Number of Factors of 69)/2} = 69^{4/2} = 69^{2}

**Ques 26** : Find the product of factors of 70.

**Solution** : Prime Factorization of 70 is 70 = 2^{1} × 5^{1 }× 7^{1}

Then, Number of Factors of 70 is (1 + 1) × (1 + 1) × (1 + 1) = 8

Thus, the product of all factors of 70 is 70^{(Number of Factors of 70)/2} = 70^{8/2} = 70^{4}

**Ques 27** : Find the product of factors of 72.

**Solution** : Prime Factorization of 72 is 72 = 2^{3} × 3^{2}

Then, Number of Factors of 72 is (3 + 1) × (2 + 1) = 12

Thus, the product of all factors of 72 is 72^{(Number of Factors of 72)/2} = 72^{12/2} = 72^{6}

**Ques 28** : Find the product of factors of 75.

**Solution** : Prime Factorization of 75 is 75 = 3^{1} × 5^{2}

Then, Number of Factors of 75 is (1 + 1) × (2 + 1) = 6

Thus, the product of all factors of 75 is 75^{(Number of Factors of 75)/2} = 75^{6/2} = 75^{3}

**Ques 29** : Find the product of factors of 81.

**Solution** : Prime Factorization of 81 is 81 = 3^{4}.

Then, Number of Factors of 81 is (4 + 1) = 5

Thus, the product of all factors of 81 is 81^{(Number of Factors of 81)/2} = 81^{5/2} = (9^{2})^{5/2} = 9^{5}

**Ques 30** : Find the product of factors of 84.

**Solution** : Prime Factorization of 84 is 84 = 2^{2} × 3^{1 }× 7^{1}

Then, Number of Factors of 84 is (2 + 1) × (1 + 1) × (1 + 1) = 12

Thus, the product of all factors of 84 is 84^{(Number of Factors of 84)/2} = 84^{12/2} = 84^{6}

**Ques 31** : Find the product of factors of 100.

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

Then, Number of Factors of 100 is (2 + 1) × (2 + 1) = 9

Thus, the product of all factors of 100 is 100^{(Number of Factors of 100)/2} = 100^{9/2} = (10^{2})^{9/2} = 10^{9}

**Ques 32** : Find the product of factors of 105.

**Solution** : Prime Factorization of 105 is 105 = 3^{1} × 5^{1 }× 7^{1}

Then, Number of Factors of 105 is (1 + 1) × (1 + 1) × (1 + 1) = 8

Thus, the product of all factors of 105 is 105^{(Number of Factors of 105)/2} = 105^{8/2} = 105^{4}

**Ques 33** : Find the product of factors of 120.

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

Then, Number of Factors of 120 is (3 + 1) × (1 + 1) × (1 + 1) = 16

Thus, the product of all factors of 120 is 120^{(Number of Factors of 120)/2} = 120^{16/2} = 120^{8}

**Ques 34** : Find the product of factors of 210.

**Solution** : Prime Factorization of 210 is 210 = 2^{1} × 3^{1 }× 5^{1 }× 7^{1}

Then, Number of Factors of 210 is (1 + 1) × (1 + 1) × (1 + 1) × (1 + 1) = 16

Thus, the product of all factors of 210 is 210^{(Number of Factors of 210)/2} = 210^{16/2} = 210^{8}

**Ques 35** : Find the product of factors of 232.

**Solution** : Prime Factorization of 232 is 232 = 2^{3} × 29^{1}

Then, Number of Factors of 232 is (3 + 1) × (1 + 1) = 8

Thus, the product of all factors of 232 is 232^{(Number of Factors of 232)/2} = 232^{8/2} = 232^{4}

**Ques 36** : Find the product of factors of 240.

**Solution** : Prime Factorization of 240 is 240 = 2^{4} × 3^{1 }× 5^{1}

Then, Number of Factors of 240 is (4 + 1) × (1 + 1) × (1 + 1) = 20

Thus, the product of all factors of 240 is 240^{(Number of Factors of 240)/2} = 240^{20/2} = 240^{10}

**Ques 37** : Find the product of factors of 360.

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

Then, Number of Factors of 360 is (3 + 1) × (2 + 1) × (1 + 1) = 24

Thus, the product of all factors of 360 is 360^{(Number of Factors of 360)/2} = 360^{24/2} = 360^{12}

**Ques 38** : Find the product of factors of 520.

**Solution** : Prime Factorization of 520 is 520 = 2^{3} × 5^{1 }× 13^{1}

Then, Number of Factors of 520 is (3 + 1) × (1 + 1) × (1 + 1) = 16

Thus, the product of all factors of 520 is 520^{(Number of Factors of 520)/2} = 520^{16/2} = 520^{8}

**Ques 39** : Find the product of factors of 2^{10}.

**Solution** : Prime Factorization of 1024 is 1024 = 2^{10 }

Then, Number of Factors of 1024 is (10 + 1) = 11

Thus, the product of all factors of 1024 is 1024^{(Number of Factors of 1024)/2} = 1024^{11/2} = (2^{10})^{11/2} = (2^{9})^{11} = 2^{99}

**Ques 40** : Find the product of factors of 1080.

**Solution** : Prime Factorization of 1080 is 1080 = 2^{3} × 3^{3 }× 5^{1}

Then, Number of Factors of 1080 is (3 + 1) × (3 + 1) × (1 + 1) = 32

Thus, the product of all factors of 1080 is 1080^{(Number of Factors of 1080)/2} = 1080^{32/2} = 1080^{16}

**Ques 41** : Find the product of factors of 1200.

**Solution** : Prime Factorization of 1200 is 1200 = 2^{4} × 3^{1 }× 5^{2}

Then, Number of Factors of 1200 is (4 + 1) × (1 + 1) × (2 + 1) = 30

Thus, the product of all factors of 1200 is 1200^{(Number of Factors of 1200)/2} = 1200^{30/2} = 1200^{15}

**Ques 42** : Find the product of factors of 1950.

**Solution** : Prime Factorization of 1950 is 1950 = 2^{1} × 3^{1 }× 5^{2 }× 13^{1}

Then, Number of Factors of 1950 is (1 + 1) × (1 + 1) × (2 + 1) × (1 + 1) = 24

Thus, the product of all factors of 1950 is 1950^{(Number of Factors of 1950)/2} = 1950^{24/2} = 1950^{12}

**Ques 43** : Find the product of factors of 2120.

**Solution** : Prime Factorization of 2120 is 2120 = 2^{3} × 5^{1 }× 53^{1}

Then, Number of Factors of 2120 is (3 + 1) × (1 + 1) × (1 + 1) = 16

Thus, the product of all factors of 2120 is 2120^{(Number of Factors of 2120)/2} = 2120^{16/2} = 2120^{8}

**Ques 43** : Find the product of factors of 2322.

**Solution** : Prime Factorization of 2322 is 2322 = 2^{1} × 3^{3 }× 43^{1}

Then, Number of Factors of 2322 is (1 + 1) × (3 + 1) × (1 + 1) = 16

Thus, the product of all factors of 2322 is 2322^{(Number of Factors of 2322)/2} = 2322^{16/2} = 2322^{8}

**Ques 44** : Find the product of factors of 3600.

**Solution** : Prime Factorization of 3600 is 3600 = 2^{4} × 3^{2 }× 5^{2}

Then, Number of Factors of 3600 is (4 + 1) × (2 + 1) × (2 + 1) = 45

Thus, the product of all factors of 3600 is 3600^{(Number of Factors of 3600)/2} = 3600^{45/2} = (60^{2})^{45/2} = 60^{45}

**Ques 45** : Find the product of factors of 18^{3}.

**Solution** : Prime Factorization of 18^{3} is 18^{3} = 2^{3} × 3^{6}

Then, Number of Factors of 18^{3} is (3 + 1) × (6 + 1) = 28

Thus, the product of all factors of 18^{3} is (18^{3})^{(Number of Factors of 18^3)/2} = 18^{3×28/2} = 18^{42}

**Ques 46** : Find the product of factors of 7056.

**Solution** : Prime Factorization of 7056 is 7056 = 2^{4} × 3^{2 }× 7^{2}

Then, Number of Factors of 7056 is (4 + 1) × (2 + 1) × (2 + 1) = 45

Thus, the product of all factors of 7056 is 7056^{(Number of Factors of 7056)/2} = 7056^{45/2} = (84^{2})^{45/2} = 84^{45}

**Ques 47** : Find the product of factors of 12 factorial.

**Solution** : Since 12! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 × 11 × 12 = 2^{10} × 3^{5} × 5^{2} × 7^{1} × 11^{1}

Then, Number of Factors of 12! is (10 + 1) × (5 + 1) × (2 + 1) × (1 + 1) × (1 + 1) = 792

Thus, the product of all factors of 12! is 12!^{(Number of Factors of 12!)/2} = (12!)^{792/2} = (12!)^{396}

**Ques 48** : Find the product of factors of 3^{11}.

**Solution** : Number of Factors of 3^{11} is (11 + 1) = 12

Thus, the product of all factors of 3^{11} is (3^{11})^{(Number of Factors of 3^11)/2} = (3^{11})^{12/2} = (3^{11})^{6} = 3^{66}

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 |