Hey guys, good day,
While prime factoring Greg got three 2s and one 3 from 24, while two 2s and three 3s from 108. But why did he combine the both info (like (from 24) three 2s + one two (from 108)) to get four 2s so that by taking sqrt root he can get two 2s? Did not get really. I know my question is confusing but this is how Greg explained
n^2 must be divisible by both 108 and 24 meaning at minimum it will be 216 {LCM of 108 and 24} and 216 = 2^3 x 3^3, so , n can be \sqrt{2^3 \times3^3} but this value under the square is not an integer ! and we want it to be divisible by both n and n^2 , so we add a 2 and a 3 to \sqrt{2^3 \times3^3} = \sqrt{2^4 \times3^4} : Now, this is divisible by both n and n^2 , simplifying \sqrt{2^4 \times3^4} = 2^2 \times 3^2 = 36
Now find the divisor of 36 :[ 1, 2, 3, 4, 6, 9, 12, 18, 36] :{12 ,36} are in our options so select it
