How many factors greater than 1 do 120, 210, and 270 have in common?
(A) One
(B) Three
(C) Six
(D) Seven
(E) Thirty
I know the answer my approach was similar to the solution given in the book find out the factors for all the given number and then find out the common. How can we solve by just making use of prime numbers?
Prime factorise the numbers:
120 = 2³·3·5
210 = 2·3·5·7
270 = 2·3³·5
And then take the minimum exponent of each factor,
(2^1)(3^1)(5^1)(7^0) or simply (2^1)(3^1)(5^1)
Now from these common factors, we can choose any possible number of combinations:
(2^1)(3^1)(5^1),
(2^0)(3^1)(5^1),
(2^1)(3^0)(5^1)…
if you know permutation and combination then you would understand that the total number of combinations would be 2x2x2 = 8 (because for each factor x there can only be 2 possibilities: (x^1) or (x^0))
But one of these possibilities is not allowed that is (2^0)(3^0)(5^0) = 1
Therefore, the answer would be (D) 7
Thanks @vidishas99!