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

It is currently 04 May 2016, 16:17
GMAT Club Tests

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide 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]

Show Tags

New post 11 Jun 2011, 17:00
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

67% (01:57) correct 33% (00:00) wrong based on 3 sessions

HideShow timer Statictics

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
1 KUDOS received
Director
Director
User avatar
Affiliations: GMATQuantum
Joined: 19 Apr 2009
Posts: 564
Followers: 103

Kudos [?]: 377 [1] , given: 9

Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink]

Show Tags

New post 11 Jun 2011, 18:21
1
This post received
KUDOS
Expert Post
GMAT Tutor
avatar
Joined: 24 Jun 2008
Posts: 1181
Followers: 368

Kudos [?]: 1206 [0], given: 4

Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink]

Show Tags

New post 11 Jun 2011, 19:46
Expert's post
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.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Director
Director
User avatar
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 650
Followers: 37

Kudos [?]: 595 [0], given: 39

Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink]

Show Tags

New post 12 Jun 2011, 04:10
The excellent solution is given in video solution. see it.
_________________

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]

Show Tags

New post 12 Jun 2011, 08:21
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: 1353
Followers: 16

Kudos [?]: 201 [0], given: 10

Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink]

Show Tags

New post 13 Jun 2011, 04:27
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, 04:27
    Similar topics Author Replies Last post
Similar
Topics:
DS Dilema: Both sufficient BUT product different answer? chartdude 0 13 Sep 2015, 13:01
1 Quantitative review 2nd edition : DS 76. Is x between 0 and Baten80 5 17 Jun 2011, 11:41
ds practice questions mastaliyev 1 05 Oct 2010, 04:52
53 Experts publish their posts in the topic Collection of 12 DS questions Bunuel 78 17 Oct 2009, 18:45
74 Experts publish their posts in the topic Collection of 8 DS questions Bunuel 50 13 Oct 2009, 20:16
Display posts from previous: Sort by

Q170 DS GMAT Review 12th Edition - Questionable answer

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| 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®.