Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club

 It is currently 29 Mar 2017, 00:49

### 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

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 y > 0, is (y^3 – y) divisible by 4?

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

### Hide Tags

Manager
Joined: 16 Feb 2010
Posts: 225
Followers: 2

Kudos [?]: 298 [0], given: 16

If y > 0, is (y^3 – y) divisible by 4? [#permalink]

### Show Tags

26 Aug 2010, 05:31
2
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

55% (02:28) correct 45% (01:47) wrong based on 186 sessions

### HideShow timer Statistics

If y > 0, is (y^3 – y) divisible by 4?

(1) y^2 + y is divisible by 10.
(2) For a certain integer k, y = 2k + 1.
[Reveal] Spoiler: OA
Manager
Joined: 16 Feb 2010
Posts: 225
Followers: 2

Kudos [?]: 298 [0], given: 16

Re: GrockIt: divisibility [#permalink]

### Show Tags

26 Aug 2010, 05:34
zisis wrote:
If y > 0, is (y3 – y) divisible by 4?

(1) y2 + y is divisible by 10.

(2) For a certain integer k, y = 2k + 1.

my method

$$1. y(y+1)$$ is div by 10

therefore if y is div by 10 (y^3 -y) is div by 4 BUT if y+1 is div by 10 therefore y= 9, 19 etc thus y+1 is not div by 4

2. y is odd therefore is not div by 4

any better methods?
Math Expert
Joined: 02 Sep 2009
Posts: 37661
Followers: 7411

Kudos [?]: 99871 [2] , given: 11060

If y > 0, is (y^3 – y) divisible by 4? [#permalink]

### Show Tags

26 Aug 2010, 07:32
2
KUDOS
Expert's post
3
This post was
BOOKMARKED
If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Not sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formating for formulas.
_________________
Manager
Joined: 24 Dec 2009
Posts: 224
Followers: 2

Kudos [?]: 41 [0], given: 3

Re: GrockIt: divisibility [#permalink]

### Show Tags

26 Aug 2010, 12:18
B for me. I believe this problem can also be solved by using prime boxes.
Manager
Joined: 21 Feb 2010
Posts: 212
Followers: 2

Kudos [?]: 32 [0], given: 1

Re: GrockIt: divisibility [#permalink]

### Show Tags

26 Aug 2010, 12:30
the key point to this question is to find out if y is an odd#...i pick B
Senior Manager
Joined: 14 Jun 2010
Posts: 331
Followers: 2

Kudos [?]: 24 [0], given: 7

Re: GrockIt: divisibility [#permalink]

### Show Tags

27 Aug 2010, 10:03
I pick B and I agree with tt11234...The key here is to pick that y is odd!
Manager
Joined: 27 May 2012
Posts: 228
Followers: 2

Kudos [?]: 72 [0], given: 446

Re: GrockIt: divisibility [#permalink]

### Show Tags

03 Mar 2014, 04:57
Bunuel wrote:
If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formating for formulas.

answer is definitely B , as the product of 2 consecutive even numbers is divisible by 4.
There is a small typo above , it should be (y-1) and (y+1) are 2 consecutive even numbers and not (y-1 ) and ( y-1) are 2 consecutive even numbers ,

Hope this prevents any confusion.
Thank you.
_________________

- Stne

Manager
Joined: 13 Apr 2015
Posts: 71
Concentration: General Management, Strategy
GMAT 1: 600 Q45 V28
GPA: 3.25
WE: Project Management (Energy and Utilities)
Followers: 1

Kudos [?]: 12 [1] , given: 325

Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]

### Show Tags

10 Nov 2015, 09:02
1
KUDOS
Bunuel wrote:
If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Not sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formatting for formulas.

I think one more case is to be considered, i.e when k = 0, y = 1. Even then 0/4 = 0 a integer. B will remain the answer
Director
Joined: 22 Jun 2014
Posts: 651
Location: United States
Concentration: General Management, Technology
Schools: IIMA , IIMB, ISB
GMAT 1: 540 Q45 V20
GPA: 2.49
WE: Information Technology (Computer Software)
Followers: 14

Kudos [?]: 244 [0], given: 95

If y is an integer greater than 0, is (y^3 - y) divisible by 4? [#permalink]

### Show Tags

15 Jan 2017, 08:20
If $$y$$ is an integer greater than 0, is ($$y^3$$ - $$y$$) divisible by 4?

(1) $$y^2$$ + $$y$$ is divisible by 10.

(2) For a certain integer $$k$$, $$y$$ = 2$$k$$ + 1
_________________

---------------------------------------------------------------
Target - 720-740
helpful post means press '+1' for Kudos!
http://gmatclub.com/forum/information-on-new-gmat-esr-report-beta-221111.html
http://gmatclub.com/forum/list-of-one-year-full-time-mba-programs-222103.html

Math Expert
Joined: 02 Sep 2009
Posts: 37661
Followers: 7411

Kudos [?]: 99871 [0], given: 11060

Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]

### Show Tags

15 Jan 2017, 08:34
HKD1710 wrote:
If $$y$$ is an integer greater than 0, is ($$y^3$$ - $$y$$) divisible by 4?

(1) $$y^2$$ + $$y$$ is divisible by 10.

(2) For a certain integer $$k$$, $$y$$ = 2$$k$$ + 1

Merging topics. Please refer to the discussion above.
_________________
Intern
Joined: 10 Jun 2016
Posts: 41
Schools: IIM
Followers: 0

Kudos [?]: 2 [0], given: 78

Re: If y > 0, is (y^3 – y) divisible by 4? [#permalink]

### Show Tags

09 Mar 2017, 11:04
Bunuel wrote:
If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Not sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formating for formulas.

Hello Bunnel or GMATCLUB members,

For Statement 2 we say since y is odd means y-1 is even and y+1 is even, so the given statement is div by 4. But consider if y = 1 since it is given y>0.
Now y-1 is 0 and y+1 = 2. Could this be a valid case ? Then the answer will be C. Request your help?
_________________

Thank You Very Much,
CoolKl
Success is the Journey from Knowing to Doing

A Kudo is a gesture, to express the effort helped. Thanks for your Kudos.

Intern
Joined: 10 Jun 2016
Posts: 41
Schools: IIM
Followers: 0

Kudos [?]: 2 [0], given: 78

If y > 0, is (y^3 – y) divisible by 4? [#permalink]

### Show Tags

09 Mar 2017, 11:07
coolkl wrote:
Bunuel wrote:
If y>0, is y^3-y divisible by 4?

$$y^3-y=y(y^2-1)=y(y-1)(y+1)$$

(1) y^2+y is divisible by 10 --> if $$y=10$$ then $$y(y-1)(y+1)=10*9*11$$, so not divisible by 4 (product of 2 odd and not multiple of 4 will not be divisible by 4) BUT if $$y=100$$ then $$y(y-1)(y+1)=100*99*101$$, so divisible by 4 (100 is a multiple of 4, hence product will be divisible by 4). Not sufficient.

(2) For a certain integer k, y=2k+1 --> $$y=odd$$ --> $$y-1=even$$ and $$y+1=even$$ --> $$y(y-1)(y+1)$$ is a multiple of 4 (product of 2 even numbers is a multiple of 4, OR as $$y-1$$ and $$y-1$$ are consecutive even numbers, thus one of them must be multiple of 4). Sufficient.

Pleas use "^" for powers, and formating for formulas.

Hello Bunnel or GMATCLUB members,

For Statement 2 we say since y is odd means y-1 is even and y+1 is even, so the given statement is div by 4. But consider if y = 1 since it is given y>0.
Now y-1 is 0 and y+1 = 2. Could this be a valid case ? Then the answer will be C. Request your help?

I got my answer, if y =1 then (y-1)y(y+1) is 0 and 0 div by 4 is still 0. This is good question.
_________________

Thank You Very Much,
CoolKl
Success is the Journey from Knowing to Doing

A Kudo is a gesture, to express the effort helped. Thanks for your Kudos.

If y > 0, is (y^3 – y) divisible by 4?   [#permalink] 09 Mar 2017, 11:07
Similar topics Replies Last post
Similar
Topics:
5 Is |x^7|*y^3*z^4>0 6 02 Feb 2015, 07:21
4 If y is an integer, is y^3 divisible by 8? 4 16 Sep 2012, 09:45
4 If y is an integer is y^3 divisible by 9 10 24 May 2011, 07:07
1 If y is an integer, is y^3 divisible by 9? 9 08 Jul 2011, 02:04
1 If y>0 is x>y? 1. 5x-4y=3 2. 4y-5x=3 2 17 Jul 2010, 08:21
Display posts from previous: Sort by

# If y > 0, is (y^3 – y) divisible by 4?

 new topic post reply 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®.