GCD of 478, 486, 888 Calculator

Given Input numbers are 478, 486, 888

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

Divisors of 478

List of positive integer divisors of 478 that divides 478 without a remainder.

1, 2, 239, 478

Divisors of 486

List of positive integer divisors of 486 that divides 486 without a remainder.

1, 2, 3, 6, 9, 18, 27, 54, 81, 162, 243, 486

Divisors of 888

List of positive integer divisors of 888 that divides 888 without a remainder.

1, 2, 3, 4, 6, 8, 12, 24, 37, 74, 111, 148, 222, 296, 444, 888

Greatest Common Divisior

We found the divisors of 478, 486, 888 . The biggest common divisior number is the GCD number.
So the Greatest Common Divisior 478, 486, 888 is 2.

Therefore, GCD of numbers 478, 486, 888 is 2

Given Input Data is 478, 486, 888

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

Prime Factorization of 478 is 2 x 239

Prime Factorization of 486 is 2 x 3 x 3 x 3 x 3 x 3

Prime Factorization of 888 is 2 x 2 x 2 x 3 x 37

Highest common occurrences in the given inputs are 2¹

Multiplying them we get the GCD as 2

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(478, 486) = 116154

GCD(478, 486) = ( 478 x 486 ) / 116154

GCD(478, 486) = 232308 / 116154

GCD(478, 486) = 2

Step2:

Here we consider the GCD from the above i.e. 2 as first number and the next as 888

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

LCM(2, 888) = 888

GCD(2, 888) = ( 2 x 888 ) / 888

GCD(2, 888) = 1776 / 888

GCD(2, 888) = 2

GCD of 478, 486, 888 is 2

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1. What is the GCD of 478, 486, 888?

GCD of 478, 486, 888 is 2

2. Where do I get the detailed procedure to find GCD of 478, 486, 888?

You can find a detailed procedure to find GCD of 478, 486, 888 on our page.

3. How to find GCD of 478, 486, 888 on a calculator?

You can find the GCD of 478, 486, 888 by simply giving the inputs separated by commas and click on the calculate button to avail the Greatest Common Divisor in less time.