swerve wrote: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.

Statement 1: x=12u, where u is an integer and x=8y+12.
In other words, x is a multiple of 12.
For x to be a multiple of 12, 8y must be a multiple of 12.

If y=3, then x = 8*3 + 12 = 36.
The GCD of 3 and 36 is 3.

If y=6, then x = 8*6 + 12 = 60.
The GCD of 6 and 60 is 6.

Since the GCD can be different values, INSUFFICIENT.

Statement 2: y=12z, where z is an integer and x=8y+12.
In other words, y is a multiple of 12.
Since we're looking for the GCD, view x in terms of its FACTORS.

If y=12, then x = 8(12) + 12 = 12(8+1) = 12*9.
The GCD of 12 and 12*9 is 12.

If y=24, then x = 8(24) + 12 = 12(8*2 + 1) = 12*17.
The GCD of 24 and 12*17 is 12.

I'm almost convinced: the GCD is 12.
Maybe one more just to be sure:

If y=36, then x = 8(36) + 12 = 12(8*3 + 1) = 12*25.
The GCD of 36 and 12*25 is 12.
SUFFICIENT.

The correct answer is B.

swerve wrote: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.
Source: GMAT Prep

\[\left\{ \begin{gathered}
x,y\,\, \geqslant \,\,1\,\,{\text{ints}} \hfill \\
x - 8y = 12\,\,\,\,\,\left( * \right) \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,?\,\, = \,\,GCD\left( {x,y} \right)\]
\[\left( 1 \right)\,\,\,x = 12u\,\,,\,\,\,u\,\,\operatorname{int} \,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,8y = 12\left( {u - 1} \right)\]
\[\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,u = 3\,\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,\,y = 3\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {x,y} \right) = \left( {12 \cdot 3\,,\,3} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 3 \hfill \\
\,{\text{Take}}\,\,u = 5\,\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,\,\,y = 6\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {x,y} \right) = \left( {12 \cdot 5\,,\,6} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 6\,\, \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,\,y = 12z\,\,,\,\,\,z\,\,\operatorname{int} \,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,x = 12 + 8 \cdot 12 \cdot z = 12\left( {8z + 1} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,? = 12\]
\[\left( {**} \right)\,\,\,GCD\,\,\left( {z\,,\,8z + 1} \right) = \,\,k \geqslant 1\,\,\,{\text{int}}\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}
\,\frac{z}{k} = {\text{in}}{{\text{t}}_{\text{1}}} \hfill \\
\,\frac{{8z + 1}}{k} = {\operatorname{int} _2}\,\,\,\,\, \hfill \\
\end{gathered} \right. \Rightarrow \,\,\,\,\,\,\,\,\frac{1}{k} = {\operatorname{int} _2} - 8\left( {\frac{z}{k}} \right) = {\operatorname{int} _2} - 8 \cdot {\operatorname{int} _1} = \operatorname{int} \,\,\,\,\,\,\,\mathop \Rightarrow \limits^{k\, \geqslant \,1\,\,\,{\text{int}}} \,\,\,\,\,k = 1\]

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

Regards,
Fabio.