Archit143 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.

\(2 < x < y\,\,{\text{ints}}\)

\(y\,\,\mathop = \limits^? \,\,16 = {2^4}\)

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

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