Can you please explain this Bunuel?

3(u-1)=2y

So, we know that 3 is a factor of 2y, but how can we conclude that 3 is a factor of y? Is it because 2 and 3 don't have any factor in common, except 1?

Exactly. 2y/3 = integer (because u-1 is an integer). This to be true, y must be a multiple of 3.
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x = 8y + 12

(1) x = 12u, where u is an integer.

If x is a multiple of 12, it means 8y is a multiple of 12. Since 8 already has three 2s, we NEED y to have a 3 but it COULD have 2s and/or other factors too.
So we cannot say what the GCD of x and y is.
Not sufficient.

(2) y = 12z, where z is an integer.
If y is a multiple of 12, x is a multiple of 12 too. So they certainly have 12 common. Let's see what else they could have common.
x is 12 more than a multiple of y so the only common factors they could have are the factors of 12. We already know that they both have 12 in them. So GCD must be 12.
(Check out this post for a discussion on this: https://anglesandarguments.com/blog-details/55)
Sufficient.

Answer (B)
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\(\left\{ \begin{gathered}
x,y\,\, \geqslant \,\,1\,\,{\text{ints}} \hfill \\
x - 8y = 12\,\,\,\,\,\left( * \right) \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,?\,\, = \,\,GCD\left( {x,y} \right)\)


\(\left( 1 \right)\,\,\,x = 12u\,\,,\,\,\,u\,\,\operatorname{int} \,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,8y = 12\left( {u - 1} \right)\)

\(\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,u = 3\,\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,\,y = 3\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {x,y} \right) = \left( {12 \cdot 3\,,\,3} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 3 \hfill \\
\,{\text{Take}}\,\,u = 5\,\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,\,y = 6\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {x,y} \right) = \left( {12 \cdot 5\,,\,6} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 6\,\, \hfill \\
\end{gathered} \right.\)


\(\left( 2 \right)\,\,\,y = 12z\,\,,\,\,\,z\,\,\operatorname{int} \,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,x = 12 + 8 \cdot 12 \cdot z = 12\left( {8z + 1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,? = 12\)

\(\left( {**} \right)\,\,\,GCD\,\,\left( {z\,,\,8z + 1} \right) = \,\,k \geqslant 1\,\,\,{\text{int}}\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}
\,\frac{z}{k} = {\text{in}}{{\text{t}}_{\text{1}}} \hfill \\
\,\frac{{8z + 1}}{k} = {\operatorname{int} _2}\,\,\,\,\, \hfill \\
\end{gathered} \right. \Rightarrow \,\,\,\,\,\,\,\,\frac{1}{k} = {\operatorname{int} _2} - 8\left( {\frac{z}{k}} \right) = {\operatorname{int} _2} - 8 \cdot {\operatorname{int} _1} = \operatorname{int} \,\,\,\,\,\,\,\mathop \Rightarrow \limits^{k\, \geqslant \,1\,\,\,{\text{int}}} \,\,\,\,\,k = 1\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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One of the hardest official DS I have seen. I have to plug in numbers to confirm things.

(1) ->

8*1 + 12 = 20
8*2 + 12 = 28
8*3 + 12 = 36 (possible)
8*4 + 12 = 44
8*5 + 12 = 52
8*6 + 12 = 60 (possible)

GCD will be the value of y and different values are possible. Insufficient.

(2) ->

96*1 + 12 = 108
96*2 + 12 = 204
96*3 + 12 = 300

For any value of z, x and y will both be divisible by 12. Sufficient.

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BrentGMATPrepNow ThatDudeKnows gmatprep ScottTargetTestPrep could you kindly share your insights on this question please? Thanks for your time in advanced.

Kimberly77

Before diving in to the question, I'll point out that while this is an official question, it's an older official question. You might find one or two questions that are this complex in the most recent OGs or Quant add-ons, but keep in mind that to get a 50, you're only expected to get half the questions of that level correct. In my opinion, the biggest takeaway here is that if you can't figure out how to wrap your head around the theory, you likely have to fall back on just trying numbers to see if you can make sense of things that way. And assuming that's the path you need to take on this question, spotting that there are two points at which we are looking at linear equations makes that job much easier. The bottom line is that when you're testing numbers, make the effort to try to spot whether there's a pattern that makes that job a little easier.

If you're trying to learn the theory, Bunuel did a good job both giving the rule and explaining how that rule is derived. I'll assume that the theory isn't clicking and you're trying to go about things by testing numbers.

Statement (1) alone:
x=12 doesn't work because that makes y=0 and we need y to be a positive integer.
x=24 doesn't work because that makes y=1.5 and we need y to be a positive integer.
x = 8y + 12 is a linear function. For each 12 that we add to x, we add 1.5 to y.
x=36, y=3 ... GCD=3.
x=48, y=4.5 ... doesn't work...y not an integer.
x=60, y=6 ... GCD=6.
We already have a GCD=3 and a GCD=6.
Does statement (1) alone give us enough information to find the GCD? No. BCE.

Statement (2) alone:
Plug y=12z into x=8y+12 to get x=8(12z)+12
x=12(8z+1)
We can see that 12 will always be a CD for x and y.
Is there a way to find a CD for z and 8z+1? (This is another linear progression where fore each 1 that we add to z, we add 8 to 8z+1.)
If z=1, 8z+1 = 9.
If z=2, 8z+1 = 17.
If z=3, 8z+1 = 25.
If z=4, 8z+1 = 33.
If z=5, 8z+1 = 41.
If z=6, 8z+1 = 49.
If z=7, 8z+1 = 57.
If z=8, 8z+1 = 65.
Never going to happen. 12 is the largest we will get.
Does statement (2) alone give us enough information to find the GCD? Yes. B.

Answer choice B.
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Brilliant and thanks ThatDudeKnows, make much more sense now.
Regarding statement 2, how will I know to spot that "We can see that 12 will always be a CD for x and y." in below and to swap 12 with 8?

Plug y=12z into x=8y+12 to get
x=8(12z)+12
x=12(8z+1)