Hello All,

My take on this question:

As given , x = 8y+12 and x and y are positive integers.

Statement 1: x = 12u, u is integer

It means x is multiple of 12. But x is also 8y+12 then it means 8y should also be multiple of x, which in turn means y can take values 3 , 6, 9 ...etc i.e. multiple of 3.
Case 1: say y = 3 then x = 8*3 + 12 = 36 , GCD(x,y) = 3
Case 2: say y = 6 then x = 8*6 +12 = 60, GCD(x,y) = 6

Clearly 1 insufficient.

Statement 2: y = 12u, u is integer

It means y is multiple of 12 , which in turn means x = 8(multiple of 12) + 12 i.e. x is nothing but the next multiple of 12 than y or y and x are consecutive multiples of 12.
So GCD of two consecutive multiples of 12 should be 12 ,hence sufficient. Answer B.

Hope it helps.

Thanks
Manoj Parashar

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?

VeritasKarishma JeffTargetTestPrep please share your detail approach to this problem. Thanks

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


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

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


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

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


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

Regards,
Fabio.
_________________

Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version)
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