Is x^2 - y^2 divisible by 8? : GMAT Data Sufficiency (DS) - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 11:11

### 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

# Is x^2 - y^2 divisible by 8?

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

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 36540
Followers: 7072

Kudos [?]: 93010 [0], given: 10541

Re: divisible by 8? [#permalink]

### Show Tags

15 Jun 2014, 10:42
mahendru1992 wrote:
Bunuel wrote:
Is $$x^2 - y^2$$ divisible by 8?

(1) $$x$$ and $$y$$ are even integers. Clearly insufficient, consider $$x=y=0$$ for an YES answer and $$x=2$$ and $$y=0$$ for a NO answer.

(2) $$x + y$$ is divisible by $$8$$ --> $$x^2 - y^2=(x+y)(x-y)$$, if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that $$x$$ and $$y$$ are integers. If $$x=4.8$$ and $$y=3.2$$, $$x+y$$ is divisible by $$8$$, BUT $$x^2 - y^2$$ is not. Not sufficient.

(1)+(2) $$x$$ and $$y$$ integers. $$x+y$$ divisible by 8. Hence $$(x+y)(x-y)$$ is divisible by $$8$$. Sufficient.

I don't know, I think I'm missing something here. If x=4.8 and y=3.2, Still the product would be divisible by 8 IMO.
(x+y)(x-y)/8
(4.8+3.2)(4.8-3.2)/8
8(1.6)/8
=1.6
What am I doing wrong here?

1.6 is not an integer. Divisible means that you get an integer as a result of division.
_________________
Intern
Joined: 10 Apr 2012
Posts: 47
Concentration: Finance
WE: Analyst (Commercial Banking)
Followers: 0

Kudos [?]: 21 [0], given: 13

Re: divisible by 8? [#permalink]

### Show Tags

16 Jun 2014, 06:34
mahendru1992 wrote:
Bunuel wrote:
Is $$x^2 - y^2$$ divisible by 8?

(1) $$x$$ and $$y$$ are even integers. Clearly insufficient, consider $$x=y=0$$ for an YES answer and $$x=2$$ and $$y=0$$ for a NO answer.

(2) $$x + y$$ is divisible by $$8$$ --> $$x^2 - y^2=(x+y)(x-y)$$, if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that $$x$$ and $$y$$ are integers. If $$x=4.8$$ and $$y=3.2$$, $$x+y$$ is divisible by $$8$$, BUT $$x^2 - y^2$$ is not. Not sufficient.

(1)+(2) $$x$$ and $$y$$ integers. $$x+y$$ divisible by 8. Hence $$(x+y)(x-y)$$ is divisible by $$8$$. Sufficient.

I don't know, I think I'm missing something here. If x=4.8 and y=3.2, Still the product would be divisible by 8 IMO.
(x+y)(x-y)/8
(4.8+3.2)(4.8-3.2)/8
8(1.6)/8
=1.6
What am I doing wrong here?

Here is where you are wrong mate:

8(1.6)/8=1.6. This clearly shows that 8(1.6) is not divisible by 8 because the result is not an integer result. Revisit your definition of divisibility. If 8 is a divisor of a number 'a', it means that 8 can get into 'a' without leaving a remainder. In other words, the operation of dividing 'a' by 8 should give an integer result, which in your case above, it isn't.

Hope that helps.
Director
Joined: 23 Jan 2013
Posts: 579
Schools: Cambridge'16
Followers: 1

Kudos [?]: 42 [0], given: 40

Re: Is x^2 - y^2 divisible by 8? [#permalink]

### Show Tags

18 Jun 2014, 22:42
mahendru1992 wrote:
Bunuel wrote:
Is $$x^2 - y^2$$ divisible by 8?

(1) $$x$$ and $$y$$ are even integers. Clearly insufficient, consider $$x=y=0$$ for an YES answer and $$x=2$$ and $$y=0$$ for a NO answer.

(2) $$x + y$$ is divisible by $$8$$ --> $$x^2 - y^2=(x+y)(x-y)$$, if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that $$x$$ and $$y$$ are integers. If $$x=4.8$$ and $$y=3.2$$, $$x+y$$ is divisible by $$8$$, BUT $$x^2 - y^2$$ is not. Not sufficient.

(1)+(2) $$x$$ and $$y$$ integers. $$x+y$$ divisible by 8. Hence $$(x+y)(x-y)$$ is divisible by $$8$$. Sufficient.

I don't know, I think I'm missing something here. If x=4.8 and y=3.2, Still the product would be divisible by 8 IMO.
(x+y)(x-y)/8
(4.8+3.2)(4.8-3.2)/8
8(1.6)/8
=1.6
What am I doing wrong here?

We get the formula (x-y)(x+y) = x^2+xy-yx-y^2=x^2-y^2, so for xy and yx it is no matter whether x and y are integer or not, they eliminate each other. Problem with squares, if x and y integers their squares will be integer, if they are not integer squares will not be integers and, what is more, their difference will never be the integer or 0. So, it will not be divisible by integer
Intern
Joined: 24 May 2013
Posts: 29
Concentration: Operations, General Management
WE: Engineering (Energy and Utilities)
Followers: 2

Kudos [?]: 14 [0], given: 21

Re: Is x^2 - y^2 divisible by 8? [#permalink]

### Show Tags

20 Jun 2014, 06:13
Bunuel wrote:
Is $$x^2 - y^2$$ divisible by 8?

(1) $$x$$ and $$y$$ are even integers. Clearly insufficient, consider $$x=y=0$$ for an YES answer and $$x=2$$ and $$y=0$$ for a NO answer.

(2) $$x + y$$ is divisible by $$8$$ --> $$x^2 - y^2=(x+y)(x-y)$$, if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that $$x$$ and $$y$$ are integers. If $$x=4.8$$ and $$y=3.2$$, $$x+y$$ is divisible by $$8$$, BUT $$x^2 - y^2$$ is not. Not sufficient.

(1)+(2) $$x$$ and $$y$$ integers. $$x+y$$ divisible by 8. Hence $$(x+y)(x-y)$$ is divisible by $$8$$. Sufficient.

Hi Bunuel,
I agree with the answer but pls tell me " if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that x and y are integers. If x=4.8 and y=3.2, x+y is divisible by 8".
Sometimes we take variable as decimal and sometimes not , i have seen questions which solve by considering x and y as integers only.
Math Expert
Joined: 02 Sep 2009
Posts: 36540
Followers: 7072

Kudos [?]: 93010 [0], given: 10541

Re: Is x^2 - y^2 divisible by 8? [#permalink]

### Show Tags

20 Jun 2014, 06:36
nidhi12 wrote:
Bunuel wrote:
Is $$x^2 - y^2$$ divisible by 8?

(1) $$x$$ and $$y$$ are even integers. Clearly insufficient, consider $$x=y=0$$ for an YES answer and $$x=2$$ and $$y=0$$ for a NO answer.

(2) $$x + y$$ is divisible by $$8$$ --> $$x^2 - y^2=(x+y)(x-y)$$, if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that $$x$$ and $$y$$ are integers. If $$x=4.8$$ and $$y=3.2$$, $$x+y$$ is divisible by $$8$$, BUT $$x^2 - y^2$$ is not. Not sufficient.

(1)+(2) $$x$$ and $$y$$ integers. $$x+y$$ divisible by 8. Hence $$(x+y)(x-y)$$ is divisible by $$8$$. Sufficient.

Hi Bunuel,
I agree with the answer but pls tell me " if one of the multiples is divisible by 8 then so is the product: true for integers, but we are not told that x and y are integers. If x=4.8 and y=3.2, x+y is divisible by 8".
Sometimes we take variable as decimal and sometimes not , i have seen questions which solve by considering x and y as integers only.

When we are told that variables are integers, then we should consider integers only. When we are not told that, then we should not assume that the variables are integers only.

By the way this question is not a good example of GMAT questions because every GMAT divisibility question will tell you in advance that any unknowns represent positive integers (ALL GMAT divisibility questions are limited to positive integers only).
_________________
Re: Is x^2 - y^2 divisible by 8?   [#permalink] 20 Jun 2014, 06:36

Go to page   Previous    1   2   [ 25 posts ]

Similar topics Replies Last post
Similar
Topics:
14 8xy^3 + 8x^3y = 2x^2*y^2 / 2^(-3), What is xy? 10 23 Feb 2012, 19:27
2 8x y^3 + 8x^3 y = 2x^2y^2 * 8 , What is xy? (1) y > x (2) 4 31 Mar 2011, 11:00
7 Is x^2 - y^2 divisible by 8? 1. x and y are even integers. 18 07 Dec 2009, 02:15
15 Is x^2+y^2 divisible by 5? 13 14 Nov 2009, 17:56
2 8x*y^3 + 8x^3*y = 2x^2*y^2 / 2^(-3), What is xy? 16 07 Oct 2007, 23:37
Display posts from previous: Sort by

# Is x^2 - y^2 divisible by 8?

 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®.