GCF of 16 and n is 4=>n must be of the form 2*2*k (k not equal to 2)
GCF of 45 and n is 3=> n must be of form 3*m (m not equal to 3 or 5)

=> n must have 2*2*3*c (c not 2, 3 , 5)

GCF of 2*2*3*c and 3*7*5*2 must have 2*3 (multiple of 6)

only 30 and 42 possible, but since c cannot be 5, GCF must be 42

hence, D

nice explanation. tx much.

The greatest common factor of 16 and the positive integer n is 4, and the greatest common factor of n and 45 is 3. Which of the following could be the greatest common factor of n and 210?

If the GCF of 16 and n is 4 the maximum number of 2s that n can have is 2
(4=2^2)
If the GCF of n and 45 is 3 the maximum number of 3s that n can have is 1

so n = k*2*2*3

Lets take n and 210 = 3*7*2*5 so we know that n and 210 will share a 2 and a 3 but not a 5 ( ? , because of the 2nd condition ) , but n and 210 can share a 7 (? , conditions 1 and 2 doesn't stop us from taking 7 as a factor of n )

So n can be 3*2*7 = 42

Answer = D

There is another way to solve this

That is by eliminating all the wrong answers

3 : cannot be 3 because we know that n has 2 as a factor
14 : cannot be 14 , because n and 210 will have 3 as a common factor
30 : Cannot be 30 because , n doesn't have 5 as a factor
42 : correct
70 : Cannot be , because n , doesn't have 5 as a factor