The answer is B.

Can someone please walk me through this problem? Specifically, how are you able to arrive at the GCF and what is it?

I know this involves finding the prime factor, but isn't the GCF dependent on the value of z, in statement 2?

Thanks!

This is a VERY difficult problem. The vast, vast majority of people will not be able to tackle this problem in 2 minutes and will need to make an educated guess and move on.

Statement (1) tells us that 12 is a divisor of x. What does it tell us about y?
12u = 8y + 12
(multiple of 12) = 8y + (multiple of 12)
8y must be a multiple of 12.

Therefore, y must be a multiple of 3 and 3 is a divisor of y. 3 might be the greatest common divisor of x and y. But y might have other divisors too (e.g., 6 or 12). Insufficient.

Statement (2) tells us that 12 is a divisor of y.
What does it tell us about x?
x = 8(12z) + 12
x = (multiple of 12) + (multiple of 12)
x must be a multiple of 12.
12 is a divisor of x.

So 12 is a common divisor of x and y. But is it the greatest common divisor?

RULE: If one number is b units away from another number, and b is a factor of both numbers, the greatest common factor of the two numbers is b. (If you want to really understand this, then think about why. Otherwise, just remember the rule.)

x (one number) is 12 units away from 8y (another number). 12 is a factor of x and 8y. Therefore, 12 is the GCF of x and 8y. The GCF of x and y can’t be bigger than the GCF of x and 8y. Thus, we can be assured that 12 is the GCF of x and y. Statement (2) alone is SUFFICIENT.

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

Thank you. Great, clear explanation.

I knew the rule, but did not know how to apply it to this problem, until now.

one way of looking at that rule is

let x-y = k an integer

if x and y are both divisible by (x-y) ... is x-y the gcf ...

then x/(x-y) = k where k is an integer. Now the question is can there be any additional divisor for k that will be common between x and y

x = xk -yk
y * k = x * (k-1)

there will be no common divisors between k and k-1 ... consecutive numbers ...

so k is indeed the largest number that can divide both x and y

this sort of approach works, yes.
for my readers, though, let me tell you that there is NO WAY the current incarnation of the gmat is going to require this sort of arithmetical gymnastics. it's no coincidence that the above question was purged after the 10th edition of the OG.

recent gmat problems are pretty much ALL focused on manipulating elementary algebra, arithmetic, etc. in unconventional and extremely clever ways.

if you have this sort of advanced background, where you can turn out number-theory approaches such as the one above, then go for it.
if not, don't worry about it.

WHERE IS THE QUESTION IN THIS !! POST ?

click the small graphic icon in the top post (it's blue and white), and you should be able to see a full-screen graphic of the problem. if you still don't see it, we can reproduce it here - let us know.

by the way, it's now forbidden to reproduce questions as images, unless the problem can't be reproduced by typing (e.g., if there are graphics).

great explanations, guys.

Here's another way to solve this problem, and it's just another useful rule, which is showed below:

*If two different natural numbers are divided by their GCD respectively, their quotients will be mutual prime numbers, which have only one factor, 1, in common.

In statement (2), as Stacey said, X = 8(12z)+12=12(8z+1); Y=12z; let's do the division and the quotients will be:

Quotient of X: 8z+1
Quotient of Y: z

It's as clear as the results shown above, so the two quotients must be mutual exclusive and have only one factor, 1, in common.

So 12 is the rght answer.

HTH!

nice work!

_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT

I'm not sure if this is correct, can someone please verify:

For Statement (2), we have:

y = 12Z

X = 8 x 12Z + 12 (substituting back into the given equation)

X = 12 (8Z + 1) --> Therefore, X is even

x = 12 x ODD

Given Y = 12Z

Is Z odd?

8 x 12Z = X -12 (using our substituted equation)

12 Z = (X - 12) / 8

Therefore Z = (X - 12) / 96 --> Even / Even = E (Since we know X is even)

Hence Z is even. X and Y therefore don't share other common factors (one is Even --> Z and the other Odd). So 12 is the GCD.

livin, wow. Nice. I can't find any flaws in your logic.

My own quick and dirty way of doing this problem would be to plug numbers.

For statement 1, I could say x=36, y=3, GCD=3.
Or x=60, y=6, GCD 6. I now have two possibilities for GCD, so insufficient.

For statement 2, I tried three sets of numbers:
y=12, x=108, GCD=12
y=24, x=204, GCD=12
y=36, x=300, GCD=12

At this point I'd feel fairly comfortable concluding 2 is sufficient. However, I would hope that this result would help me recall the theory Stacey mentioned and then I could confidently conclude that the answer is B.

_________________
Jamie Nelson
ManhattanGMAT Instructor

thanks for that quick method jamie, that's very helpful.

Great to hear! Thank you again for your prior post--it added a lot to this conversation.

_________________
Jamie Nelson
ManhattanGMAT Instructor

can someone tell me how to know 8z+1 and z are mutual and only have common factor .1 . THANK YOU!