Greatest Common Factor of 147, 510, 717, 912, 357

Created By : Jatin Gogia

Reviewed By : Rajasekhar Valipishetty

Last Updated : Apr 06, 2023


GCF of two or more numbers Calculator allows you to quickly calculate the GCF of 147, 510, 717, 912, 357 i.e. 3 largest integer that divides all the numbers equally.

Greatest common factor (GCF) of 147, 510, 717, 912, 357 is 3.

GCF(147, 510, 717, 912, 357) = 3

GCF of 147, 510, 717, 912, 357

Greatest common factor or Greatest common divisor (GCD) can be calculated in two ways

GCF of:

Greatest Common Factor of 147,510,717,912,357

GCF of 147,510,717,912,357 is 3

3 147, 510, 717, 912, 357
49, 170, 239, 304, 119

∴ So the GCF of the given numbers is 3 = 3

Greatest Common Factor (GCF) By Matching Biggest Common Factor Method

Factors of 147

List of positive integer factors of 147 that divides 147 without a remainder.

1,3,7,21,49,147

Factors of 510

List of positive integer factors of 510 that divides 510 without a remainder.

1,2,3,5,6,10,15,17,30,34,51,85,102,170,255,510

Factors of 717

List of positive integer factors of 717 that divides 717 without a remainder.

1,3,239,717

Factors of 912

List of positive integer factors of 912 that divides 912 without a remainder.

1,2,3,4,6,8,12,16,19,24,38,48,57,76,114,152,228,304,456,912

Factors of 357

List of positive integer factors of 357 that divides 357 without a remainder.

1,3,7,17,21,51,119,357

Greatest Common Factor

We found the factors 147,510,717,912,357 . The biggest common factor number is the GCF number.
So the greatest common factor 147,510,717,912,357 is 3.

GCF of two or more Numbers Calculation Examples

Frequently Asked Questions on GCF of 147, 510, 717, 912, 357

1. What is the GCF of 147, 510, 717, 912, 357?

Answer: GCF of 147, 510, 717, 912, 357 is 3.

2. How to Find the GCF of 147, 510, 717, 912, 357

Answer: Greatest Common Factor(GCF) of 147, 510, 717, 912, 357 = 3

Step 1: Divide all the numbers with common prime numbers having remainder zero.

Step 2: Then multiply all the prime factors GCF(147, 510, 717, 912, 357) = 3.