If x and y are integers and 2 < x < y, does y = 16 ?

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If x and y are integers and 2 < x < y, does y = 16 ?

(1) The GCF of X and Y is 2.
(2) The LCM of X and Y is 48.

OA C

Source: Veritas Prep

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BTGmoderatorDC wrote:If x and y are integers and 2 < x < y, does y = 16 ?

(1) The GCF of X and Y is 2.
(2) The LCM of X and Y is 48.
Source: Veritas Prep
\[2 < x < y\,\,{\text{ints}}\]
\[y\,\,\mathop = \limits^? \,\,16 = {2^4}\]

\[\left( 1 \right)\,\,GCF\left( {x,y} \right) = 2\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {{2^2},2 \cdot 3} \right)\,\,\, \Rightarrow \,\,\,\left\langle {{\text{No}}} \right\rangle \hfill \\
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {2 \cdot {{3,2}^4}} \right)\,\,\, \Rightarrow \,\,\,\left\langle {{\text{Yes}}} \right\rangle \hfill \\
\end{gathered} \right.\]

\[\left( 2 \right)\,\,LCM\left( {x,y} \right) = {2^4} \cdot 3\,\,\,\left\{ \begin{gathered}
\,\left( {\operatorname{Re} } \right){\text{Take}}\,\,\left( {x,y} \right) = \left( {2 \cdot {{3,2}^4}} \right)\,\,\, \Rightarrow \,\,\,\left\langle {{\text{Yes}}} \right\rangle \hfill \\
\,{\text{Take}}\,\,\left( {x,y} \right) = \left( {{2^2}{{,2}^4} \cdot 3} \right)\,\,\, \Rightarrow \,\,\,\left\langle {{\text{No}}} \right\rangle \hfill \\
\end{gathered} \right.\]

\[\left( {1 + 2} \right)\,\,\,\left\{ \begin{gathered}
\,\left( 1 \right) \cap \left( 2 \right)\,\,\,\, \Rightarrow \,\,\,\,xy = GCF\left( {x,y} \right) \cdot LCM\left( {x,y} \right) = 3 \cdot {2^5}\,\,\,\left( * \right) \hfill \\
\,\left( 1 \right)\,\, \Rightarrow \left\{ \begin{gathered}
x = 2 \cdot M \hfill \\
y = 2 \cdot N \hfill \\
\end{gathered} \right.\,\,\,\,\,GCF\left( {M,N} \right) = 1\,\,,\,\,\left( * \right)\,\, \Rightarrow \,\,3 \cdot {2^3} = MN\,\,\,\,\left( {M,N \geqslant 2} \right) \hfill \\
\left( {M,N} \right) = \left( {{2^3},3} \right)\,\,\, \Rightarrow \,\,\,\left( {x,y} \right) = \left( {{2^4},2 \cdot 3} \right)\,\,\, \Rightarrow \,\,\,x > y\,\,{\text{impossible}} \hfill \\
\therefore \left( {M,N} \right) = \left( {{{3,2}^3}} \right)\,\,\, \Rightarrow \,\,\,\left( {x,y} \right) = \left( {2 \cdot {{3,2}^4}} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\text{Yes}}} \right\rangle \hfill \\
\end{gathered} \right.\]

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

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Jay@ManhattanReview » Mon Nov 26, 2018 9:09 pm

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BTGmoderatorDC wrote:If x and y are integers and 2 < x < y, does y = 16 ?

(1) The GCF of X and Y is 2.
(2) The LCM of X and Y is 48.

OA C

Source: Veritas Prep
Let's take each statement one by one.

(1) The GCF of X and Y is 2.

Case 1: Say x = 2 and y = 16; we have GCF of x and y = 2, and y = 16. The answer is Yes.
Case 2: Say x = 2 and y = 32; we have GCF of x and y = 2, and y ≠ 16. The answer is No.

No unique answer.Insufficient.

(2) The LCM of X and Y is 48.

Case 1: Say x = 3 and y = 16; we have LCM of x and y = 48, and y = 16. The answer is Yes.
Case 2: Say x = 16 and y = 24; we have LCM of x and y = 48, and y ≠ 16. The answer is No.

No unique answer.Insufficient.

(1) and (2) together

From Statement (1), say x = 2p and y = 3q, where p and q are co-prime and p < q

Note that the product of two numbers is equal to the product of their GCF and LCM.

=> xy = 2*48

=> 4pq = 2*48

=> pq = 24

=> pq = 24 = 1*24 = 2*12 = 3*8 = 4*6;

We see that only for 24 = 3*8, we have 3 and 8 are co-prime to each other; thus, p = 3 and q = 8.

So, y = 2q = 2*8 = 16.

The answer is yes. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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