How do you find the greatest common factor of 88, 66?

##GCF (88, 66) = 22##, but there is another way to calculate it, sometimes more usefull…

Calculate the ##GCF## making a list of the factors of numbers and looking for the biggest of those who are repeated is a simple method but it can be very slow to use if we have more than two numbers and they are of a large size.

Instead, using the other method I describe below, you can calculate the ##GCF## fairly quickly, whatever numbers we have to consider, and the strategy used also serves to other operations and related integer calculations (eg , calculating the ##LCM##, simplifying radicals or fractions …).

(1) For each of the numbers that we have to consider, we make its :

##color(white) “0000”##For example, suppose you want to find the ##GCF## of ##600##, ##1500## ##color(white) “0000”##and ##3300##. The factorization of these three numbers is:

##color(white) “00000000000000000000” 600 = 2^3 cdot 3 cdot 5^2## ##color(white) “0000000000000000000” 1500 = 2^2 cdot 3 cdot 5^3## ##color(white) “0000000000000000000” 3300 = 2^2 cdot 3 cdot 5^2 cdot 11##

(2) We chose those factors that are repeated in all numbers, first taking the base of each.

##color(white) “0000”##In our example, as the powers with equal bases on the three ##color(white) “0000”##numbers are those with base 2, 3 and 5, those would be the ##color(white) “0000”##factors to consider. The factor 11, however, only appears in the ##color(white) “0000”##decomposition of one of the numbers, so we discard it:

##color(white) “000000000000” GCF (600, 1500, 3300) = 2^? cdot 3^? cdot 5^?##

(3) We must use as , for each base, the smallest of which appear in the prime factorization.

##color(white) “0000”##Of all the factors that have ##2## as a base, the smallest exponent ##color(white) “0000”##that appears is the ##2##, therefore, we will use ##2^2## in calculating the ##color(white) “0000” GCF##. We do the same with the ##3## (which is raised to ##1## in the ##color(white) “0000”##three numbers, so we’ll use ##3^1##) and ##5## (which has the smallest ##color(white) “0000”##exponent ##2##):

##color(white) “000000000000” GCF (600, 1500, 3300) = 2^2 cdot 3 cdot 5^2 = 300##.

We can recall the method of calculating the ##GCF## learning that “we take only those factors that are repeated, and using the smallest possible exponent”. More abbreviated form:

##color(white) “0000” GCF = “common factors with lower exponent”##.