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 ::
GMATH method creator (Math for the GMAT)
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