Find all School-related info fast with the new School-Specific MBA Forum

It is currently 20 Aug 2014, 14:37

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Q170 DS GMAT Review 12th Edition - Questionable answer

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 11 Jun 2011
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink] New post 11 Jun 2011, 16:00
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

50% (02:40) correct 50% (00:00) wrong based on 2 sessions
Dear All,

I just need someone to please take a look at this question in GMAT Review 12th Edition, DS, Q170.

If n is a positive integer, is n3 - n divisible by 4?

(1) n=2k + 1, where k is an integer,
(2) n2 + n is divisible by 6.

GMAT answer is A.

However.

n3 - n = n (n2 - 1) = n ( n+1) (n-1) = (n-1) n (n+1), i.e. 3 cons. integers.

n = 2K + 1, where k is an integer. n is a positive ODD integer, i.e. 2K + 1 > 0 and K> -1/2.

So, we can have cons. integers:

0, 1, 2 (where zero is an even integer)

2, 3, 4
and so on, where we are going to have at least one "four", therefore the answer seems to be sufficient EXCEPT we have this possibility of 0, 1, 2 where, as we know, zero is an even integer. n = 2K + 1 holds and so does (n-1) n (n+1). So, in my opinion, this is INSUFFICIENT.

Please help.

Thank you.

Alex
[Reveal] Spoiler: OA
Kaplan Promo CodeKnewton GMAT Discount CodesGMAT Pill GMAT Discount Codes
1 KUDOS received
Manager
Manager
User avatar
Joined: 19 Apr 2009
Posts: 170
Location: San Francisco, California
Followers: 36

Kudos [?]: 166 [1] , given: 1

Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink] New post 11 Jun 2011, 17:21
1
This post received
KUDOS
Here is my video solution.

http://www.gmatquantum.com/list-of-vide ... ds170.html

Dabral
_________________

Free Video Explanations: OFFICIAL GUIDE GMAT 13, 12, 11, 10; QUANT REVIEW 2nd, 1st.

GMAT Instructor
avatar
Joined: 24 Jun 2008
Posts: 967
Location: Toronto
Followers: 252

Kudos [?]: 651 [0], given: 3

GMAT Tests User
Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink] New post 11 Jun 2011, 18:46
Aleks1974 wrote:
Dear All,

I just need someone to please take a look at this question in GMAT Review 12th Edition, DS, Q170.

If n is a positive integer, is n3 - n divisible by 4?

(1) n=2k + 1, where k is an integer,
(2) n2 + n is divisible by 6.

GMAT answer is A.

However.

n3 - n = n (n2 - 1) = n ( n+1) (n-1) = (n-1) n (n+1), i.e. 3 cons. integers.

n = 2K + 1, where k is an integer. n is a positive ODD integer, i.e. 2K + 1 > 0 and K> -1/2.

So, we can have cons. integers:

0, 1, 2 (where zero is an even integer)

2, 3, 4
and so on, where we are going to have at least one "four", therefore the answer seems to be sufficient EXCEPT we have this possibility of 0, 1, 2 where, as we know, zero is an even integer. n = 2K + 1 holds and so does (n-1) n (n+1). So, in my opinion, this is INSUFFICIENT.


Zero is divisible by every positive integer. Remember we say that a is divisible by b if a/b is an integer. So zero is divisible by 4, for example, because 0/4 = 0, which is an integer. So the case where n^3 - n = 0 is not an exception in this question; in that case n^3 - n is certainly divisible by 4.
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Private GMAT Tutor based in Toronto

Director
Director
User avatar
Status: GMAT Learner
Joined: 14 Jul 2010
Posts: 652
Followers: 34

Kudos [?]: 213 [0], given: 32

GMAT Tests User
Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink] New post 12 Jun 2011, 03:10
The excellent solution is given in video solution. see it.
_________________

I am student of everyone-baten
Collections:-
PSof OG solved by GC members: http://gmatclub.com/forum/collection-ps-with-solution-from-gmatclub-110005.html
DS of OG solved by GC members: http://gmatclub.com/forum/collection-ds-with-solution-from-gmatclub-110004.html
100 GMAT PREP Quantitative collection http://gmatclub.com/forum/gmat-prep-problem-collections-114358.html
Collections of work/rate problems with solutions http://gmatclub.com/forum/collections-of-work-rate-problem-with-solutions-118919.html
Mixture problems in a file with best solutions: http://gmatclub.com/forum/mixture-problems-with-best-and-easy-solutions-all-together-124644.html

Intern
Intern
avatar
Joined: 11 Jun 2011
Posts: 2
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink] New post 12 Jun 2011, 07:21
IanStewart wrote:
Aleks1974 wrote:
Dear All,

I just need someone to please take a look at this question in GMAT Review 12th Edition, DS, Q170.

If n is a positive integer, is n3 - n divisible by 4?

(1) n=2k + 1, where k is an integer,
(2) n2 + n is divisible by 6.

GMAT answer is A.

However.

n3 - n = n (n2 - 1) = n ( n+1) (n-1) = (n-1) n (n+1), i.e. 3 cons. integers.

n = 2K + 1, where k is an integer. n is a positive ODD integer, i.e. 2K + 1 > 0 and K> -1/2.

So, we can have cons. integers:

0, 1, 2 (where zero is an even integer)

2, 3, 4
and so on, where we are going to have at least one "four", therefore the answer seems to be sufficient EXCEPT we have this possibility of 0, 1, 2 where, as we know, zero is an even integer. n = 2K + 1 holds and so does (n-1) n (n+1). So, in my opinion, this is INSUFFICIENT.


Zero is divisible by every positive integer. Remember we say that a is divisible by b if a/b is an integer. So zero is divisible by 4, for example, because 0/4 = 0, which is an integer. So the case where n^3 - n = 0 is not an exception in this question; in that case n^3 - n is certainly divisible by 4.


Thank you! That's what I needed.

Alex
VP
VP
avatar
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1365
Followers: 11

Kudos [?]: 135 [0], given: 10

GMAT Tests User
Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink] New post 13 Jun 2011, 03:27
B is not sufficient because,
n = 2 , 3 give different values.

A is always divisible. Sufficient.
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Re: Q170 DS GMAT Review 12th Edition - Questionable answer   [#permalink] 13 Jun 2011, 03:27
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic Kaplan or GMAT official 12th edition review?? cynthiap 1 13 Dec 2011, 09:29
OG Guide 12th Ed DS Q:170 sterlinggrey 2 24 Apr 2011, 07:09
Possible Error GMAT Review 12th edition asch13 2 03 Jun 2010, 21:05
1 gmat 12th edition review book question srjonz 2 06 Dec 2009, 08:32
Qn 124 in DS Section of GMAT 12th Edition Official Review? junker 1 16 Nov 2009, 17:06
Display posts from previous: Sort by

Q170 DS GMAT Review 12th Edition - Questionable answer

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.