Is the value of 27^a/6^b a prime number? (1) 3a – b = 1 (2) b is a no

answer has to be C.
b = -1 and let a= -1/3
answer can be Prime.

Given = 27^a/6^b

which can be written as = 3^3a/2^b*3^b

= 3^3a-b/2^b

Statement 1: 3a-b = 1

Which leaves us with 3/2^b

We don't have any info about b so if b = 0, we will be left with 3 which is prime. If b =1 we will be left with 3/2.

So statement 1 is insufficient.

Statement 2: b is a non-zero integer, which in itself in not sufficient as it can have any value from -ve to +ve and we don't have any info on a.

= 3^3a-b/2^b

if we put a = -1/3 and b = -1 this will get solved to 2 which is a prime and there can be scenarios when it will not be prime. So this statement itself isn't sufficient.

Taking statement 1 and 2 together.

3^3a-b/2^b

From st 1 we know 3a-b = 1 and st 2 b is a non zero integer. So we know its not a prime.

So option C IMHO

Hi mbaprep2016

as per highlighted part if we put b=-1 in statement 1 then a=0 not 1/3

Thanks

Hi rohit8865,

The source of the question has been added. Thanks for reminding me.

Cheers!

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.