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

 It is currently 16 Sep 2014, 07:50

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If n is a positive integer, is n^3 – n divisible by 4?

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Intern
Joined: 28 Jul 2009
Posts: 2
Followers: 0

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

If n is a positive integer, is n^3 – n divisible by 4? [#permalink]  18 Aug 2009, 07:53
00:00

Difficulty:

55% (hard)

Question Stats:

68% (02:02) correct 32% (01:42) wrong based on 81 sessions
If n is a positive integer, is n^3 – n divisible by 4 ?

(1) n = 2k + 1, where k is an integer
(2) n^2 + n is divisible by 6
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 29641
Followers: 3488

Kudos [?]: 26202 [0], given: 2706

Re: GMAT Official Guide 12th Edition DS Question 170 [#permalink]  05 Sep 2009, 18:35
Expert's post
If n is a positive integer, is n^3 – n divisible by 4 ?

n^3-n=n(n^2-1)=(n-1)n(n+1), so we are asked whether the product of 3 consecutive integers is divisible bu 4.

(1) n = 2k + 1, where k is an integer --> n=odd --> as n is odd then both n-1 and n+1 are even hence (n-1)n(n+1) is divisible by 4. Sufficient.

(2) n^2 + n is divisible by 6 --> if n=2 then n^3-n=6 and the answer is NO but if n=3 then n^3-n=24 and the answer is YES. Not sufficient.

Answer: A.
_________________
SVP
Joined: 05 Jul 2006
Posts: 1542
Followers: 5

Kudos [?]: 76 [0], given: 39

Re: GMAT Official Guide 12th Edition DS Question 170 [#permalink]  06 Sep 2009, 03:14
[quote="bambazoon"]This question is from the 12th ed. of the official guide and I think there's something wrong with the solution.
I'm sure others have done this question already, but I haven't been able to find a related topic.

The question goes:
If n is a positive integer, is n^3 - n divisible by 4?

1) n = 2k + 1, where k is an integer
2) n^2 + n is divisible by 6

(n-1)(n+1)n ........ 2 odd , even or 2 even,odd

from 1
suff

from 2

(n)(n+1) = 6n..........even , odd we dont know the 3rd..........insuff

A
Intern
Joined: 20 Aug 2009
Posts: 42
Followers: 1

Kudos [?]: 8 [0], given: 5

Re: GMAT Official Guide 12th Edition DS Question 170 [#permalink]  06 Sep 2009, 10:11
Is n(n+1)(n-1) divisible by 4, n is positive
stat1: n= 2K+1 => n is odd, in that case (n-1) and (n+1) are even
=> the product should be divisible by 4 ....suff.

stat2:n(n+1) is divisible by 6 => one no. is even( multiple of 2) and other one is odd and odd no. is multiple of 3
=> the product may or may not be divisible by 4 (divisibility will hold only for even no.s those are multiple of 4)...insuff.
IMO A
Intern
Joined: 23 Nov 2011
Posts: 1
Followers: 0

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

Re: GMAT Official Guide 12th Edition DS Question 170 [#permalink]  23 Nov 2011, 13:27
Hi there, I have a question: Can I not just test cases for statement 1? Say k= 1, 2 and 5, then n would be 3, 5 and 11, respectively. Then plug into n(cube)-n/4: 3 and 5 would work, 11 wouldn't --> insufficient.
Intern
Joined: 14 Sep 2010
Posts: 24
Followers: 0

Kudos [?]: 6 [0], given: 4

Re: GMAT Official Guide 12th Edition DS Question 170 [#permalink]  14 Jan 2012, 05:53
If n is a positive integer, is n^3 - n divisible by 4?

1) n = 2k + 1, where k is an integer

2) n^2 + n is divisible by 6

Statement 1: n = 2k + 1

(2k + 1)^3 =

(4k^2 + 4k + 1)(2k + 1) =

8k^3 + (4k^2 + 8k^2) + (4k + 2k + 1).

n^3 - n is derived by subtracting
2k + 1 from the binomial.

8k^3 + (4k^2 + 8k^2) + (4k)

Each term of n^3 - n is divisible by 4 where n is odd.

Statement 2: n^2 + n is divisible by 6

If n = 2, n^2 + n = 6, and
(2)^3 - 2 = 6 is not divisible by 4.
However, if n = 3, n^2 + n = 12, and (3)^3 - 3 = 24 is divisible by 4.

Posted from my mobile device
Manager
Joined: 28 Jul 2013
Posts: 59
Location: India
Concentration: Marketing, Strategy
GPA: 3.62
WE: Engineering (Manufacturing)
Followers: 0

Kudos [?]: 8 [0], given: 20

Re: If n is a positive integer, is n^3 – n divisible by 4? [#permalink]  02 Jun 2014, 23:54
first statement says k is an integer....so if k=0 then n will be 1 and the product of n(n-1)(n+1) will be zero....in that it is not divisible by 4

pls illustrate
Math Expert
Joined: 02 Sep 2009
Posts: 29641
Followers: 3488

Kudos [?]: 26202 [0], given: 2706

Re: If n is a positive integer, is n^3 – n divisible by 4? [#permalink]  03 Jun 2014, 00:43
Expert's post
sayansarkar wrote:
first statement says k is an integer....so if k=0 then n will be 1 and the product of n(n-1)(n+1) will be zero....in that it is not divisible by 4

pls illustrate

Zero is divisible by EVERY integer except zero itself, (or, which is the same, zero is a multiple of every integer except zero itself).
_________________
Re: If n is a positive integer, is n^3 – n divisible by 4?   [#permalink] 03 Jun 2014, 00:43
Similar topics Replies Last post
Similar
Topics:
1 If n is a positive integer, is n3 n divisible by 4? 1. n = 6 24 Feb 2008, 17:59
If n is a positive integer, is n^3 - n divisible by 4? (1) n 2 07 Oct 2007, 12:20
If n is a positive integer, is n^3 - n divisible by 4? 1) n 7 27 Jul 2007, 20:55
If n is a positive integer, is n^3-n divisible by 4? 1) n = 7 15 Apr 2006, 20:07
If n is a positive integer, is n^3 - n divisible by 4? (1) n 15 24 Oct 2005, 21:12
Display posts from previous: Sort by

# If n is a positive integer, is n^3 – n divisible by 4?

 Question banks Downloads My Bookmarks Reviews Important topics

 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®.