One more GCD LCM problem

Hi Vittalgmat,

For the GCD problem ..........
Is the answer A?.

Please reply.

CarryOn

The question sounds really weird to me.

If a and b are both divisible by 6, then GCD(a,b) is indeed 6 [whatever values a and b are]. This answers the question. There is no need for data sufficiency here. Data is already sufficient.

Experts, opinions please?

but they could also be divisible by 12, 24, 48 ...



How can we surely say that 6 is GCD ?

vittalgmat wrote:If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?

(1) a = 2b + 6

(2) a = 3b

[spoiler]OA is probably A (not sure)[/spoiler]

If you don't recall the relevant rule or are uncomfortable with proceeding algebraically (as many test-takers might be here), you can handle this question fairly quickly by picking numbers.

We know from the stem that both a and b are positive integers divisible by 6; in other words, 6 is a factor of both a and b. The question is asking: Is 6 their greatest common factor?

(1) a = 2b + 6

So, let's say b is 6. Then, a = 2(6) + 6 = 18. Is 6 the greatest common factor of 6 and 18? Yes. But we need to convince ourselves that a and b will always have 6 as their greatest common factor. So, now, let's say b is 12. Then, a = 30. Is 6 the greatest common factor of 12 and 30? Again, yes. (And, if b = 18, then a = 42. Is 6 the greatest common factor of 18 and 42? Yes: 6 is the biggest number that can divide both 18 and 42.) You can go for another trial if you want but, at this point, we have convinced ourselves that no matter what multiples of 6 we let a and b be, 6 will always be their greatest common factor.

Sufficient.


(2) a = 3b

If b = 6, then, a = 18. Is 6 the greatest common factor of 6 and 18? Yes. But if b = 12, then a = 36. Is 6 the greatest common factor of 12 and 36? No; 12 is. Because, from (2), we can get both a yes and a no answer to the yes/no question, this statement is insufficient.

The first statement is sufficient by itself while the second one is not; choice A.