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

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

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Updated on: 26 Oct 2016, 00:14
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Question Stats:

49% (02:22) correct 51% (01:36) wrong based on 67 sessions

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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.

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Originally posted by duahsolo on 25 Oct 2016, 16:22.
Last edited by duahsolo on 26 Oct 2016, 00:14, edited 1 time in total.
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Joined: 28 Jun 2016
Posts: 207
Concentration: Operations, Entrepreneurship
Re: Is the value of 27^a/6^b a prime number? (1) 3a – b = 1 (2) b is a no  [#permalink]

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25 Oct 2016, 16:54
duahsolo wrote:
Is the value of 27^a/6^b a prime number?

(1) 3a – b = 1
(2) b is a non-zero integer.

Statement 1:

If b=0, then a=1/3 and 27^a/6^b = 3 ---------------Yes

If b=1, Then a=2/3 and 27^a/6^b = 9/6 = 3/2 -----------No

Insufficient

Statement 2:

If b\neq{0} and b is an integer.

Then 27^a/6^b can never be prime number

Sufficient

IMO B
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Re: Is the value of 27^a/6^b a prime number? (1) 3a – b = 1 (2) b is a no  [#permalink]

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25 Oct 2016, 21:32
duahsolo wrote:
Is the value of 27^a/6^b a prime number?

(1) 3a – b = 1
(2) b is a non-zero integer.

Hi duahsolo

Could u please specify the source of your questions/check OA(last 5-6 posts) ???

Per (2) if b>0 then odd/even=non integer---->Not Prime
if b<0 then odd*even=even (2 is only even prime,which is never possible as b is integer)----->Not Prime

suff

Ans B

thanks
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Re: Is the value of 27^a/6^b a prime number? (1) 3a – b = 1 (2) b is a no  [#permalink]

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25 Oct 2016, 21:54
1
b = -1 and let a= -1/3
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Posts: 110
Re: Is the value of 27^a/6^b a prime number? (1) 3a – b = 1 (2) b is a no  [#permalink]

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25 Oct 2016, 22:16
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
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Re: Is the value of 27^a/6^b a prime number? (1) 3a – b = 1 (2) b is a no  [#permalink]

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25 Oct 2016, 23:01
mbaprep2016 wrote:
b = -1 and let a= -1/3

Hi mbaprep2016

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

Thanks
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Re: Is the value of 27^a/6^b a prime number? (1) 3a – b = 1 (2) b is a no  [#permalink]

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26 Oct 2016, 00:17
rohit8865 wrote:
duahsolo wrote:
Is the value of 27^a/6^b a prime number?

(1) 3a – b = 1
(2) b is a non-zero integer.

Hi duahsolo

Could u please specify the source of your questions/check OA(last 5-6 posts) ???

Per (2) if b>0 then odd/even=non integer---->Not Prime
if b<0 then odd*even=even (2 is only even prime,which is never possible as b is integer)----->Not Prime

suff

Ans B

thanks

Hi rohit8865,

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

Cheers!
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Re: Is the value of 27^a/6^b a prime number? (1) 3a – b = 1 (2) b is a no  [#permalink]

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26 Sep 2018, 15:10
1
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.
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Re: Is the value of 27^a/6^b a prime number? (1) 3a – b = 1 (2) b is a no &nbs [#permalink] 26 Sep 2018, 15:10
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