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

It is currently 17 Sep 2014, 01:51

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

Is x^2+y^2 divisible by 5?

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Manager
Manager
avatar
Joined: 24 Jul 2009
Posts: 76
Location: United States
GMAT 1: 590 Q48 V24
Followers: 2

Kudos [?]: 35 [1] , given: 124

Is x^2+y^2 divisible by 5? [#permalink] New post 14 Nov 2009, 17:56
1
This post received
KUDOS
4
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

28% (01:39) correct 72% (01:23) wrong based on 41 sessions
Is x^2+y^2 divisible by 5?

(1) When x-y is divided by 5, the remainder is 1
(2) When x+y is divided by 5, the remainder is 3
Intern
Intern
avatar
Joined: 12 Nov 2009
Posts: 7
Followers: 0

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

Re: Remainder Problem, divisible by 5? [#permalink] New post 14 Nov 2009, 19:01
Hi,

I'm not sure if this is correct, but my answer is D. Both put together can be used to answer the question.

Explanation: Since 1 says x-y gives a remainder of 1 when divided by 5, this gives an option that x-y has to end in either 3 or 8. 2 says that x+y gives a remainder of 3 when divided by 5. Hence x+y has to end in either 1 or 6. If we try the possible combinations for x and y then it can only be when they end with 2 and 1 respectively. So if we square up numbers ending with 2 and 1 the unit digits will be 4 and 1 and hence sum to 5 which is divisible by 5.

Hence x^2+y^2 is divisible by 5.

Please correct me if I am wrong
VP
VP
avatar
Joined: 05 Mar 2008
Posts: 1477
Followers: 11

Kudos [?]: 195 [0], given: 31

GMAT Tests User
Re: Remainder Problem, divisible by 5? [#permalink] New post 14 Nov 2009, 19:07
avinarvind wrote:
Hi,

I'm not sure if this is correct, but my answer is D. Both put together can be used to answer the question.

Explanation: Since 1 says x-y gives a remainder of 1 when divided by 5, this gives an option that x-y has to end in either 3 or 8. 2 says that x+y gives a remainder of 3 when divided by 5. Hence x+y has to end in either 1 or 6. If we try the possible combinations for x and y then it can only be when they end with 2 and 1 respectively. So if we square up numbers ending with 2 and 1 the unit digits will be 4 and 1 and hence sum to 5 which is divisible by 5.

Hence x^2+y^2 is divisible by 5.

Please correct me if I am wrong


I'm getting answer E

1. x-y/5 gives remainder 1

therefore x-y=1, 6, 11, 16, 21
insufficient because we don't know x or y

2. x+y/5 gives remainder 3
therefore x+y = 3, 8, 13, 18, 23
insufficient

the formula is y = qx + r

equation 1: x-y = 5q + 1
equation 2: x+y = 5q + 3

combine equations and solve
x+y = x - y - 1 + 3
2y = 2
y = 1

if y = 1 x can be 7
7+1 = 8 and divided by 5 gives remainder 3
7-1 = 6 and divided by 5 gives remainder 1

however

17 + 1 = 18 and divided by 5 gives remainder 3
17-1 = 16 and divided by 5 gives remainder 1

so answer E
Intern
Intern
avatar
Joined: 12 Nov 2009
Posts: 7
Followers: 0

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

Re: Remainder Problem, divisible by 5? [#permalink] New post 14 Nov 2009, 21:55
I am not sure if I understood the above explanation.

But from what I see, the question is whether x^2+y^2 is divsible by 5.

So as per the explanation x^2 + y^2 is divisible by 5 since y=1 and x has to end in 7. The unit digit will always be 9+1=0 and hence divisible by 5. So answer is 'D'.
Intern
Intern
avatar
Joined: 26 Sep 2009
Posts: 10
Followers: 0

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

Re: Remainder Problem, divisible by 5? [#permalink] New post 15 Nov 2009, 04:10
Is x^2+y^2 divisible by 5?
1). When x-y is divided by 5, the remainder is 1
2). When x+y is divided by 5, the remainder is 3

Answer:

Statement 1:
This tells us two things-
1) Either one of x and y is a multiple of 5
or
2) Both x and y are non -multiples of 5-( x-y=can result in multiple of 5 or non multiple of 5 )
Both x and y CANNOT be multiples of 5 -Not sufficient

Statement 2:
This tells us the same thing as 1
1) Either one of x and y is a multiple of 5
or
2) Both x and y are non -multiples of 5( x+y=can result in multiple of 5 or non multiple of 5 )
Both x and y CANNOT be multiples of 5 -Not sufficient

Combining (1) and (2)
Either one of x and y is a multiple of 5 or Both x and y are non -multiples of 5
this means
either x2 or y2 is a multiple of 5 or both are non multiples of 5
x2+y2- this can result in a multiple of 5 or a non multiple - so Not sufficient

Answer -(E)
VP
VP
avatar
Joined: 05 Mar 2008
Posts: 1477
Followers: 11

Kudos [?]: 195 [0], given: 31

GMAT Tests User
Re: Remainder Problem, divisible by 5? [#permalink] New post 15 Nov 2009, 07:12
just realized..i didn't even answer the question...I apologize..I think the answer should be d not e
Expert Post
6 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 29659
Followers: 3494

Kudos [?]: 26255 [6] , given: 2708

Re: Remainder Problem, divisible by 5? [#permalink] New post 15 Nov 2009, 07:38
6
This post received
KUDOS
Expert's post
2
This post was
BOOKMARKED
ctrlaltdel wrote:
Is x^2+y^2 divisible by 5?
1). When x-y is divided by 5, the remainder is 1
2). When x+y is divided by 5, the remainder is 3


First I must say this is not GMAT question. As every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

But let's deal with question as it's written. x^2+y^2 to be divisible by 5 it should be an integer first and then should have the last digit 0 or 5:

(1) x-y=5n+1 --> (x-y)^2=25n^2+10n+1 --> x^2+y^2=25n^2+10n+1+2xy. Now 25n^2+10n is divisible by 5, but what about 1-2xy? Not sufficient.

(2) x+y=5m+3 --> (x+y)^2=25m^2+30m+9 --> x^2+y^2=25m^2+30m+9-2xy. The same here. Not sufficient.

(1)+(2) Add the equations:
2(x^2+y^2)=25n^2+10n+25m^2+30m+10=25(n^2+m^2)+10(n+3m)+10.

x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+3m)+5. Well if n^2+m^2 is even, x^2+y^2 is an integer and it's divisible by 5. But what if n^2+m^2 is odd, then x^2+y^2 is not an integer at all, hence not divisible by 5.

Can we somehow conclude that n^2+m^2 is even? If we are not told that x and y are integers then not.

Answer: E.

To check this consider following pairs of x an y: x=4.5, y=3.5 and x=7, y=1,
x=4.5, y=3.5
x-y=4.5-3.5=1 first statement holds true, 1 divided by 5 remainder 1.
x+y=4.5+3.5=8 second statement holds true, 8 divided by 5 remainder 3.
x^2+y^2=4.5^2+3.5^2=32.5 not divisible by 5.

x=7, y=1
x-y=7-1=6 first statement holds true, 6 divided by 5 remainder 1.
x+y=7+1=8 second statement holds true, 8 divided by 5 remainder 3.
x^2+y^2=7^2+1^2=50 is divisible by 5.

Two different answers. Not sufficient. Answer: E.


Now let's see if we were told that x and y are positive integers, as GMAT does. Still would be quite hard 700+ question.

Last step:
x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+3m)+5. Can we now conclude that that n^2+m^2 is even?

x-y=5n+1 and x+y=5m+3. Add them up: 2x=5(n+m)+4 --> x=\frac{5(n+m)}{2}+2. As x is an integer n+m must be divisible by 2, hence either both are even or both are odd, in any case n^2+m^2 is even and divisible by 2. Hence x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+15m)+5 is an integer and divisible by 5.

In this case answer C.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 24 Jul 2009
Posts: 76
Location: United States
GMAT 1: 590 Q48 V24
Followers: 2

Kudos [?]: 35 [0], given: 124

Re: Remainder Problem, divisible by 5? [#permalink] New post 15 Nov 2009, 07:45
Bunuel, you rock... :thanks
Intern
Intern
avatar
Joined: 18 Nov 2009
Posts: 2
Followers: 0

Kudos [?]: 1 [0], given: 19

Re: Remainder Problem, divisible by 5? [#permalink] New post 18 Nov 2009, 19:53
That was tough, you are scaring me.
I just joined yesterday with a hope that i will get good DS questions.
This is too tough :(
Senior Manager
Senior Manager
avatar
Joined: 21 Mar 2010
Posts: 316
Followers: 5

Kudos [?]: 21 [0], given: 33

GMAT Tests User
Re: Remainder Problem, divisible by 5? [#permalink] New post 26 Feb 2011, 08:26
Hmm- im thinking the key to any remainder question is the equation! will try to implement this as a thumb rule!
Intern
Intern
avatar
Joined: 09 Jan 2012
Posts: 2
Followers: 0

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

Re: no.prop [#permalink] New post 11 Jan 2012, 01:34
1- case
(x-y)/5 --> 1 (Remainder)
(x-y)^2/5--> 1(Remainder)
nothing can be deduced for remainder of x^2+y^2 divided by 5
insufficient

2- case

(x+y)/5-->3 (remainder)
(x+y)^2/5-->4(remainder)

insufficient as nothing can be deduced for remainder of x^2+y^2 divided by 5

however adding two cases
2(x^2+y^2)/5--> remainder 0 (1+4=5 divisible by 5) hence divisible by 5

hence both the cases are required

alternatively take x=12, y=6
and cross check

hence answer is c

what is OA
Manager
Manager
avatar
Joined: 03 Oct 2009
Posts: 64
Followers: 0

Kudos [?]: 30 [0], given: 8

Re: Is x^2+y^2 divisible by 5? 1). When x-y is divided by 5, the [#permalink] New post 17 Jan 2012, 21:59
Is x^2+y^2 divisible by 5?
1). When x-y is divided by 5, the remainder is 1

Insufficient

2). When x+y is divided by 5, the remainder is 3

Insufficient


1 + 2


When (x-y)^2 is divided by 5, remainder will be 1

(x-y)^2 = (x^2+y^2 - 2xy) = 5p + 1 -> A

-----

When (x+y)^2 is divided by 5, remainder will be 4

(x+y)^2 = (x^2+y^2 + 2xy) = 5k + 4 -> B

A + B

2(x^2+y^2) = 5(p+k)+5
2(x^2+y^2) = 5(p+k+1)
(x^2+y^2) = (5/2)(p+k+1) -> for (x^2+y^2) to be integer , (p+k+1) has to be divisible by 2.

Hence, (x^2+y^2) is divisible by 5.
Intern
Intern
avatar
Joined: 05 Jun 2005
Posts: 2
Concentration: Strategy, Finance
GMAT Date: 04-05-2014
WE: Engineering (Manufacturing)
Followers: 0

Kudos [?]: 5 [0], given: 27

Re: Is x^2+y^2 divisible by 5? [#permalink] New post 27 Feb 2014, 09:26
Bunuel, you are awesome....damn.,..that is such a great explanation!!!!
Re: Is x^2+y^2 divisible by 5?   [#permalink] 27 Feb 2014, 09:26
    Similar topics Author Replies Last post
Similar
Topics:
14 Experts publish their posts in the topic Is x^2 - y^2 divisible by 8? icandy 24 09 May 2009, 10:25
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 an integer divisible by 9? 1) x is an integer MooseDrool 7 09 Aug 2007, 16:22
Is x^2+y^2 divisible by 5? 1). When x-y is divided by 5, the johnycute 7 20 Nov 2006, 05:53
Equation |x/2| + |y/2| = 5 encloses a certain region on the allabout 6 20 Dec 2005, 04:16
Display posts from previous: Sort by

Is x^2+y^2 divisible by 5?

  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®.