If M = 14^5*21^3, and N = 49^2*6^3 , then what is the number of all

Question

If  M = 14^5*21^{3} , and  N = 49^2*6^3 , then what is the number of all distinct positive factors of the greatest common factor of M and N?

(A) 3
(B) 55
(C) 70
(D) 80
(E) 90

IanStewart · Answer

Getting prime factorizations,

M = (14^5)(2^{13}) = (2^5)(7^5)(2^{13}) = (2^{18})(7^5)

and

N = (49^2)(6^3) = (7^4)(2^3)(3^3)

To find the GCD of two numbers from their prime factorizations (something I've never needed to do on a real GMAT question), we just see where the prime factorizations overlap. Looking at each prime, the GCD should include 2^3, but no higher power of 2 (because N is not divisible by 2^4 or any higher power of 2), and it should include 7^4, but no higher power of 7, and it should not include any 3's, because M is not divisible by 3. So the GCD is  (2^3)(7^4) . To count a number's divisors from its prime factorization, we add 1 to each exponent and multiply what we get, so this GCD has (3 + 1)(4 + 1) = 4*5 = 20 different divisors in total.

That's not among the answer choices, so there's something wrong with the question. I wonder if it means to say  M = (14^5)(6^{13}) , in which case we'd also end up with a 3^3 in the GCD, and thus with 80 divisors instead of 20.

Bunuel · Answer

It should have been  M = 14^5*21^{3}  instead of  M = 14^5*2^{13} . My bad. Edited. Thank you!

AnuK2222 · Answer

Solution D:80
M= 2^3*3^3*7^8
N=2^3*3^3*7^4
Greatest common factor aka Highest common factor is product of lowest powers of common prime factors= 2^3*3^3*7^4
no. of positive factors is product of +1 of power of each of the prime factors ie. (3+1)*(3+1)*(4+1)=80

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If M = 14^521^3, and N = 49^26^3 , then what is the number of all

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