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

It is currently 25 Oct 2014, 22:14

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

What is the remainder when x^2+y^2 is divided by 5, where x

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
avatar
Joined: 08 Aug 2005
Posts: 251
Followers: 1

Kudos [?]: 9 [0], given: 0

What is the remainder when x^2+y^2 is divided by 5, where x [#permalink] New post 06 May 2006, 04:31
What is the remainder when x^2+y^2 is divided by 5, where x and y are positive integer?

1). When x+y is divided by 5, the remainder is 1.
2). When x-y is divided by 5, the remainder is 2.
Manager
Manager
avatar
Joined: 09 Apr 2006
Posts: 173
Location: Somewhere in Wisconsin!
Followers: 1

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

 [#permalink] New post 07 May 2006, 10:40
(I) gives (x + y) = {1,11,21,...} or {6,16,26,...}
(II) gives (x - y) = {2,12,22,..} or {7,17,27,...}

(x^2 + y^2) = (x + y)^2 - 2xy = (x - y)^2 + 2xy

As we do not get any information about xy from either (I) or (II) we can rule out A and B.

Now combine them.

(x+y)^2(mod5) = 1 (as the last digit would be 1 or 6)
(x-y)^2(mod5) = 4 (as the last digit would be 4 or 9)

Add them and you would get 2(x^2 + y^2) on the LHS.

On the RHS you would get 1+4=5. So 2(x^2 + y^2)(mod 5) = 0 ==> (x^2 + y^2)(mod 5) = 0. So ans is C.
_________________

Thanks,
Zooroopa

Senior Manager
Senior Manager
avatar
Joined: 29 Jun 2005
Posts: 403
Followers: 1

Kudos [?]: 15 [0], given: 0

 [#permalink] New post 08 May 2006, 08:52
Agree with C
here is brutal and unneccessary approach:

From st1 we learned that x+y is a number that ends to 6 or 1
From st2 we learned that x-y ends to 7 or 2
So lets assume, that x+y=16, x-y=12. x=14, y=2. the remainder for x^2+y^2 is 0.
Now, lets assume that x+y=21, x-y=17. x=19, y=2. the remainder is 0.

from these, we can see that one of them will always end to 4 or 9, and the other to 1 or 4 respectively, which gives us 5 at the end of X^2+Y^2
Thus the remainder is always 0, and the answer is C
Intern
Intern
avatar
Joined: 03 May 2006
Posts: 27
Followers: 0

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

 [#permalink] New post 08 May 2006, 09:48
Its C. Each individual is INSUFF. Combining. Square and add.
which gives 2(x^2 + y^2) = multiple of 5.
Since x and y are integers, (x^2+y^2) is multiple of 5.
  [#permalink] 08 May 2006, 09:48
    Similar topics Author Replies Last post
Similar
Topics:
2 Experts publish their posts in the topic What is the remainder when x^2 - y^2 is divided by 3? manishuol 6 29 Apr 2013, 11:05
42 Experts publish their posts in the topic If x and y are integer, what is the remainder when x^2 + y^2 kt750 23 26 Apr 2012, 19:58
Experts publish their posts in the topic If y = ( x^2 - y^2)/(x-y) where x not equal to y, then what eyunni 6 21 Nov 2007, 11:32
Is x^2+y^2 divisible by 5? 1). When x-y is divided by 5, the goalsnr 1 10 Aug 2007, 09:53
Is x^2+y^2 divisible by 5? 1). When x-y is divided by 5, the johnycute 7 20 Nov 2006, 05:53
Display posts from previous: Sort by

What is the remainder when x^2+y^2 is divided by 5, where x

  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®.