GCD of 798, 625, 675 Calculator

Given Input numbers are 798, 625, 675

To find the GCD of numbers using factoring list out all the divisors of each number

Divisors of 798

List of positive integer divisors of 798 that divides 798 without a remainder.

1, 2, 3, 6, 7, 14, 19, 21, 38, 42, 57, 114, 133, 266, 399, 798

Divisors of 625

List of positive integer divisors of 625 that divides 625 without a remainder.

1, 5, 25, 125, 625

Divisors of 675

List of positive integer divisors of 675 that divides 675 without a remainder.

1, 3, 5, 9, 15, 25, 27, 45, 75, 135, 225, 675

Greatest Common Divisior

We found the divisors of 798, 625, 675 . The biggest common divisior number is the GCD number.
So the Greatest Common Divisior 798, 625, 675 is 1.

Therefore, GCD of numbers 798, 625, 675 is 1

Given Input Data is 798, 625, 675

Make a list of Prime Factors of all the given numbers initially

Prime Factorization of 798 is 2 x 3 x 7 x 19

Prime Factorization of 625 is 5 x 5 x 5 x 5

Prime Factorization of 675 is 3 x 3 x 3 x 5 x 5

The above numbers do not have any common prime factor. So GCD is 1

Step1:

Let's calculate the GCD of first two numbers

The formula of GCD is GCD(a, b) = ( a x b) / LCM(a, b)

LCM(798, 625) = 498750

GCD(798, 625) = ( 798 x 625 ) / 498750

GCD(798, 625) = 498750 / 498750

GCD(798, 625) = 1

Step2:

Here we consider the GCD from the above i.e. 1 as first number and the next as 675

The formula of GCD is GCD(a, b) = ( a x b) / LCM(a, b)

LCM(1, 675) = 675

GCD(1, 675) = ( 1 x 675 ) / 675

GCD(1, 675) = 675 / 675

GCD(1, 675) = 1

GCD of 798, 625, 675 is 1

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• GCD of 927, 412, 687
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1. What is the GCD of 798, 625, 675?

GCD of 798, 625, 675 is 1

2. Where do I get the detailed procedure to find GCD of 798, 625, 675?

You can find a detailed procedure to find GCD of 798, 625, 675 on our page.

3. How to find GCD of 798, 625, 675 on a calculator?

You can find the GCD of 798, 625, 675 by simply giving the inputs separated by commas and click on the calculate button to avail the Greatest Common Divisor in less time.