Factoring - Greatest Common Divisor

If $$a$$ and $$b$$ are positive integers, and $$a=2b+6$$, the greatest common divisor of $$a$$ and $$b$$ CANNOT be

Alternative explanation If 12 is the common divisor of $$a$$ and $$b$$, it means that both $$a$$ and $$b$$ are divisible by 12. The problem with 12 is that $$a$$ and $$b$$ cannot both be divisible by 12. If $$b$$ is divisible by 12, then $$a=2b+6$$ is the sum of two expressions—one of which, $$2b$$, *is* a multiple of 12, and one of which, 6, isn't. Since the sum of a multiple of 12 and a nonmultiple of 12 can never equal a multiple of 12, it follows that if $$a$$ is a multiple of 12, $$b$$ is *not* a multiple. Thus, 12 cannot be the GCD of $$a$$ and $$b$$. However, if $$b$$ is a multiple of 12, then $$2b$$ is also a multiple of 12. But since 6 is not, $$a=2b+6$$ cannot be a multiple of 12. This is, again, because the sum of a multiple of 12 and a nonmultiple of 12 *cannot* equal a multiple of 12. Thus, it is impossible for both $$a$$ and $$b$$ to be multiples of 12, so the GCD *cannot* be 12.

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Incorrect. [[snippet]] If $$b=2$$, then $$a = 2\cdot 2+6 = 10$$. The GCD of 2 and 10 is 2, so 2 *can* be the GCD of $$a$$ and $$b$$. POE 2.

Incorrect. [[snippet]] If $$b=3$$, then $$a = 2\cdot 3+6 = 12$$. The GCD of 3 and 12 is 3, so 3 *can* be the GCD of $$a$$ and $$b$$. POE 3.

Incorrect. [[snippet]] If $$b=6$$, then $$a = 2\cdot 6+6 = 18$$. The GCD of 6 and 18 is 6, so 6 *can* be the GCD of $$a$$ and $$b$$. POE 6.

Correct. [[snippet]] If $$b=2$$, then $$a = 2\cdot2+6 = 10$$. The GCD of 2 and 10 is 2, so 2 *can* be the GCD of $$a$$ and $$b$$. POE 2. If $$b=3$$, then $$a = 2\cdot 3+6 = 12$$. The GCD of 3 and 12 is 3, so 3 *can* be the GCD of $$a$$ and $$b$$. POE 3. If $$b=6$$, then $$a = 2\cdot 6+6 = 18$$. The GCD of 6 and 18 is 6, so 6 *can* be the GCD of $$a$$ and $$b$$. POE 6. Thus, the only remaining answer choice, 12, must be the right one.

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