duahsolo wrote:

Is the value of 27^a/6^b a prime number? (Source: Bell Curves)

(1) 3a – b = 1

(2) b is a non-zero integer.

Beautiful problem duahsolo. (Kudos!)

\(\frac{{{{27}^a}}}{{{6^b}}} = \frac{{{3^{3a - b}}}}{{{2^b}}}\,\,\mathop = \limits^? \,\,{\text{prime}}\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ \begin{gathered}

\,b = 0\,\,\,{\text{and}}\,\,\,3a - b = 1 \hfill \\

\,{\text{OR}} \hfill \\

b = - 1\,\,{\text{and}}\,\,3a - b = 0 \hfill \\

\end{gathered} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ \begin{gathered}

\,\left( {a\,;\,b} \right)\,\,\mathop = \limits^? \,\,\,\left( {\frac{1}{3};0} \right) \hfill \\

\,{\text{OR}} \hfill \\

\,\left( {a\,;\,b} \right)\,\,\mathop = \limits^? \,\,\,\left( { - \frac{1}{3}; - 1} \right) \hfill \\

\end{gathered} \right.\)

\(\left( 1 \right)\,\,\,3a - b = 1\,\,\,\,\,\left\{ \begin{gathered}

\,{\text{Take}}\,\,\left( {a\,;\,b} \right)\,\, = \,\,\left( {\frac{1}{3};0} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\

\,{\text{Take}}\,\,\left( {a\,;\,b} \right)\,\, = \,\,\left( {1;2} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\

\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,b \ne 0\,\,\,\operatorname{int} \,\,\,\,\,\,\left\{ \begin{gathered}

\,{\text{Take}}\,\,\left( {a\,;\,b} \right)\,\, = \,\,\left( {1;2} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\, \hfill \\

\,{\text{Take}}\,\,\left( {a\,;\,b} \right)\,\, = \,\,\left( { - \frac{1}{3}; - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\

\end{gathered} \right.\,\,\)

\(\left( {1 + 2} \right)\,\,\,\left\{ \begin{gathered}

\,\left( {a\,;\,b} \right)\,\, = \,\,\left( {\frac{1}{3};0} \right)\,\,\,\,{\text{contradicts}}\,\,\left( 2 \right)\,\, \hfill \\

\,\,\left( {a\,;\,b} \right)\,\, = \,\,\left( { - \frac{1}{3}; - 1} \right)\,\,\,\,{\text{contradicts}}\,\,\left( 1 \right)\,\, \hfill \\

\end{gathered} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle\)

The correct answer is (C), indeed.

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

Regards,

Fabio.

_________________

Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)

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