GMATPrep Practice Test #2 Question

[Guest]

If x and y are positive integers such that x=8y+12, what is the greatest common divisor of x and y?
(1) x=12u, where u is an integer
(2) y=12z, where z is an integer

This is a data sufficiency problem. Please explain!

[RonPurewal]

This is a tough problem. You should know right away to approach it with prime factorizations ("prime boxes"), though the rest is difficult.

Let's look at part 1 first. Here are some conclusions we can reach from part 1: (note that a period signifies multiplication; I couldn't use "x" for fear of creating confusion with the variable x)
* x = 2.2.3.u
* 8y + 12 = 12u
--> 2y + 3 = 3u
--> 2y = 3(u-1)
--> y = 3(u-1)/2 but still an integer.
* The only thing we know with confidence from this stuff is that y is a multiple of 3.
This doesn't look too promising - the prime box would only contain 3 and (u-1)/2, certainly not enough to reach a valid conclusion.
Other approaches don't give much of value either (and you don't want to spend the time!). Since the yield of results is so paltry, let's try plugging in particular multiples of 3 for y.
* Let y = 3. Then x = 8(3) + 12 = 36, and the greatest common divisor is 3.
* Let y = 6. Then x = 8(6) + 12 = 60, and the greatest common divisor is 6.
So (1) IS NOT SUFFICIENT.

Whew.

Now let's look at #2.
* y = 2.2.3.z (the "prime box" contains 2, 2, 3, and z)
* x = 8y + 12 = 2.2.(2y + 3)
= 2.2.(2.2.2.3.z + 3)
= 2.2.3.(8z + 1) (the "prime box" contains 2, 2, 3, and 8z+1)
But we know that z and 8z+1 don't have anything in common - any factor of z is also a factor of 8z, and therefore NOT a factor of 8z+1.
This means that the greatest common divisor is 2.2.3 = 12. So (2) IS SUFFICIENT.

Answer = B

[Guest]

Thanks so much for the reply. Wow--this indeed is a hard problem, although I would argue that we needn't need to think of the prime box to solve it. In your explanation for (1), (u-1)/2 is not necessarily prime since it could be 1. That's where the "prime box" confused me, although I understand that you are using the technique of breaking down the numbers into factors. In your explanation for (2), z should not have to be prime either... it could be 1 and I think the problem is a bit simpler if you just do: x=8y+12=8(12z)+12=12(8z+1) and use your logic about z and 8z+1.

[yamini]

what is prime boxes 2,2,3...etc?

[RonPurewal]

Guest:

Well, right, the quantities you've mentioned don't have to be prime. But we'll still put them in the box, because we can't tell whether they are prime, and so that's the best way to treat them. In the example of z, you're right that z could be 1; however, all that matters is that z and 8z+1 don't share any factors. I didn't raise the issue of z=1 because, frankly, the problem is complicated enough!

In any case: Your logic is correct, and I think we both agree on the main crux of the problem.