How to solve this in greg's way?

What this question means is you have to find the greatest common factor aka highest common factor.

Now break each number into its prime factor components.

120=2^3*3^1*5^1
210=2^1*3^1*5^1*7^1
270=2^1*3^3*5^1

Clearly the common factors for all the three numbers are 2^1 * 3^1 * 5^1…so total factors are 8 (2*2*2): adding 1 to each of the powers;
However, 1 is also included as a factor in those 8 factors and we have to exclude that because it says factors higher than 1…so remaining factors = 8-1 =7
I think the answer is D

You’re right bro the answer is 7

Hi thank you so much but if it’s possible could you please explain me from gcm (2
1

3
1

5
1
) How did we get total factors 8?

Formula for finding out total numbers of factors of a numberhttps://gmatclub.com/forum/finding-number-of-factors-of-an-integer-question-163773.html

Hey thanks a lot! I forgot the formula … Thanks a lot!