Last visit was: 25 Apr 2026, 07:27 It is currently 25 Apr 2026, 07:27
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
duahsolo
Joined: 02 Jun 2015
Last visit: 31 Jul 2023
Posts: 143
Own Kudos:
773
 [5]
Given Kudos: 1,196
Location: Ghana
Posts: 143
Kudos: 773
 [5]
1
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
User avatar
acegmat123
Joined: 28 Jun 2016
Last visit: 25 Oct 2021
Posts: 146
Own Kudos:
Given Kudos: 99
Location: Canada
Concentration: Operations, Entrepreneurship
Posts: 146
Kudos: 220
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
rohit8865
Joined: 05 Mar 2015
Last visit: 19 Apr 2026
Posts: 815
Own Kudos:
Given Kudos: 45
Products:
Posts: 815
Kudos: 1,008
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
mbaprep2016
Joined: 29 May 2016
Last visit: 30 Jun 2018
Posts: 70
Own Kudos:
101
 [1]
Given Kudos: 362
Posts: 70
Kudos: 101
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
answer has to be C.
b = -1 and let a= -1/3
answer can be Prime.
User avatar
gauravk
Joined: 29 Aug 2008
Last visit: 23 Jul 2020
Posts: 79
Own Kudos:
Given Kudos: 284
Schools: AGSM '18
Posts: 79
Kudos: 102
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
rohit8865
Joined: 05 Mar 2015
Last visit: 19 Apr 2026
Posts: 815
Own Kudos:
Given Kudos: 45
Products:
Posts: 815
Kudos: 1,008
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mbaprep2016
answer has to be C.
b = -1 and let a= -1/3
answer can be Prime.

Hi mbaprep2016

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

Thanks
User avatar
duahsolo
Joined: 02 Jun 2015
Last visit: 31 Jul 2023
Posts: 143
Own Kudos:
Given Kudos: 1,196
Location: Ghana
Posts: 143
Kudos: 773
Kudos
Add Kudos
Bookmarks
Bookmark this Post
rohit8865
duahsolo
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!
User avatar
fskilnik
Joined: 12 Oct 2010
Last visit: 03 Jan 2025
Posts: 883
Own Kudos:
1,887
 [1]
Given Kudos: 57
Status:GMATH founder
Expert
Expert reply
Posts: 883
Kudos: 1,887
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
duahsolo
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.
Moderators:
Math Expert
109827 posts
498 posts
212 posts