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If a and b are positive integers divisible by 6, is 6 the greatest com

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abhimahna

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23 Oct 2016, 09:09

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camilles

Thanks for the explanation.

Still I do not understand why statement 2 is not sufficient. My reasoning is this one:
- for 6 to be GCD(a,b) then "a" and "b" must be multiple of 6 but "a" must also be equal to b+6 right?
- The only way to have a=3b and a=b+6 is for b=3 and a=9. However as b/6 must result in an integer we know that "a" cannot be equal to 3b, hence it is enough to say that 6 is NOT the gcd of (a,b). Therefore statement 2 is sufficient to answer the question.

Can you please tell me where my reasoning is flawed?

Take b = 12, you will get a = 36.

Now try to calculate the HCF of a and b, you will get your answer, why B is insufficient.

KarishmaB

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25 Oct 2023, 22:33

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gmatbull

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

Recall that the GCD of two numbers must be a factor of their difference.
6 is a factor of a and b.

(1) a = 2b + 6

a - 2b = 6
This means that GCD of a and 2b is almost 6. Then GCD of a and b cannot be more than 6 either (because b has a subset of factors of 2b).
Since both a and b have 6 as a factor, 6 must be their GCD.
Sufficient.

(2) a = 3b
If a = 3b, we need the GCD of 3b and b. That will simply be 'b'. If b is 6, the GCD will be 6. If b is 12, the GCD will be 12 etc.
Not sufficient.

Answer (A)