How do you find the greatest common factor (GCF) of two numbers?

1 Answer
Oct 16, 2017

Here are four methods...

Explanation:

There are several methods. Which one is most convenient probably depends on the nature of the numbers.

Method 1 - Factor both numbers first

This method involves expressing each of the numbers as a product of prime numbers and identifying the common factors.

For example:

#252 = 2 * 2 * 3 * 3 * 7#

#90 = 2 * 3 * 3 * 5#

So the GCF of #252# and #90# is:

#2 * 3 * 3 = 18#

Method 2 - Integer division

Given two numbers, proceed as follows:

  • Divide the larger number by the smaller to give a quotient and remainder.

  • If the remainder is #0# then the smaller number is the GCF.

  • Otherwise repeat with the smaller number and the remainder.

So in our example...

#252 / 90 = 2" "# with remainder #72#

#90 / 72 = 1" "# with remainder #18#

#72 / 18 = 4" "# with remainder #0#

So the GCF of #252# and #90# is #18#

Method 3 - Subtraction

Given two numbers, subtract the larger from the smaller until the two numbers are the same, which is the GCF.

So with #252# and #90# it goes like this:

#252 - 90 = 162#

#162 - 90 = 72#

#90 - 72 = 18#

#72 - 18 = 54#

#54 - 18 = 36#

#36 - 18 = 18#

So the GCF of #252# and #90# is #18#

Method 4 - Simultaneous factoring

In this method we identify common factors and discard unique factors until one of the numbers becomes #1# or the numbers are identifiably coprime.

For example, given #252# and #90#...

  • Note that both are divisible by #2# since they both end in an even digit. So note down #color(red)(2)# and divide both numbers by #2# to give: #126# and #45#.

  • Note that #126# is divisible by #2# but #45# is not. So discard the unique factor #2# by dividing #126# by #2# to leave us: #63# and #45#.

  • Note that both #63# and #45# are divisible by #3# since the sum of their digits is divisible by #3#. So note down #color(red)(3)# and divide both numbers by #3# to get: #21# and #15#.

  • Note that both #21# and #15# are divisible by #3# again. So note down #color(red)(3)# and divide to get: #7# and #5#.

  • #7# and #5# are both prime and have no common factor, so we can stop and gather the factors we found:

    #color(red)(2) * color(red)(3) * color(red)(3) = color(blue)(18)#