If a and b are positive integers divisible by 6

[Pathik]

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

This is from manhattanGMAT CAT exam.

Why is A sufficient?

Tutors, could you please describe the theory below in simpler words.

"If one number is x units away from another number, and x is also a factor of both of those numbers, than x is also the GCF of those two numbers. "

[UPA]

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

This is from manhattanGMAT CAT exam.

Why is A sufficient?

Tutors, could you please describe the theory below in simpler words.

"If one number is x units away from another number, and x is also a factor of both of those numbers, than x is also the GCF of those two numbers. "

From i: a = 2b + 6
if we suppose b = 6k where k is an integer
a = 2(6k) +6
a = 6 (2k+1)

then the GCD of 6 (2k+1) and 6k is always 6. so suff....

From ii: a = 3b
suppose b = 6k where k is an integer
a = 3 (6k)
a = 18k

then the GCD of 18k and 6k is 6k. since we donot know about k, 6k could be 6 or any multiple of 6. so nsf..

So A makes sense...

[RonPurewal]

Tutors, could you please describe the theory below in simpler words.

"If one number is x units away from another number, and x is also a factor of both of those numbers, than x is also the GCF of those two numbers. "

i don't know about simpler words, but i can probably help by illustration. specifically, consider the case in which 'x' is 6.
then this means:
if you have two multiples of 6 that are only 6 apart, then their greatest common factor is 6.

this ought to make sense to you; here's a sketch of the necessary reasoning:
(step 1) the greatest common factor is AT LEAST 6:
this must be true, because 6 goes into both numbers. therefore 6 is a common factor of the two numbers. we're not done yet, though, because we haven't established that it's the greatest common factor.
(step 2) the greatest common factor is AT MOST 6:
the numbers are only 6 apart. therefore it's impossible that any number more than 6 could go into both of them, because they aren't far enough apart. (if you have two multiples of 12, for instance, they have to be at least 12 apart, and so on)

the same is true for all other integers as well; there's nothing special about 6.

[vivs.gupta]

Hi, I have one doubt regarding answer provided in the manhattan exam. For option (b), it says insufficient, with the following explaination:

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(2) INSUFFICIENT: This statement can also be tested with real numbers. If b = 6, then a would be 18 and the GCF would be 6. If, however, b = 12, then a would be 36 and the GCF would be 12.

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Now, kindly read the question:
If a and b are positive integers divisible by 6, is 6 the greatest common divisor of a and b?

Based on option (b), the answer is 'NO'. 6 is not the GCF. So, in my view option (b) is also sufficient and answer should be D.

Can you let me know your views on this?

[tim]

when you say "option (b)" i assume you mean statement 2. the information in statement 2 does not lead to a "NO" answer; it leads to a "MAYBE". this is made clear by the solution, which demonstrates sample numbers where the answer is yes and sample answers where the answer is no. this is why statement 2 is insufficient..