If x and y are integer, what is the remainder when x^2 + y^2

Oldest

Best Reply

Author

Message

TAGS:

Intern

Joined: 02 Apr 2012

Posts: 1

Followers: 0

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

If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

26 Apr 2012, 20:58

This post received
KUDOS

00:00

Question Stats:

52% (02:40) correct

47% (01:27) wrong

based on 10 sessions

If x and y are integer, what is the remainder when x^2 + y^2 is divided 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 2

[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Apr 2012, 21:56, edited 1 time in total.

Edited the question

Senior Manager

Joined: 11 May 2011

Posts: 383

Location: US

Followers: 1

Kudos [?]: 46 [0], given: 46

Re: if x and y are integers [#permalink]

26 Apr 2012, 21:11

kt750 wrote:

if x and y are integer, what is the remainder when x^2 and y^2 is divided 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 2

how do you solve this problem?

My Answer is E becasue (x^2 + y^2)/5 can't be calculated either with x-y or with x+y.

Cheers.
_________________

-----------------------------------------------------------------------------------------
What you do TODAY is important because you're exchanging a day of your life for it!
-----------------------------------------------------------------------------------------

Intern

Joined: 22 Jan 2012

Posts: 35

Followers: 0

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

Re: if x and y are integers [#permalink]

26 Apr 2012, 21:19

1) x-y= 5k+1
tells us only about the difference between x and y , not about the individual numbers. hence not sufficient.

2) x+y=5m+2
tells us only about the addition of x and y , not about the individual numbers. hence not sufficient.

1&2) adding both the conditions:
2x= 5k+5m+3
2x= 5(k+m)+3
x= 1/2(5(k+m)+3)
inserting some positive values of k,m we can find x and thereby x^2/5.

for y subtract both the conditions:
2y= 5m+2 - 5k-1
2y= 5(m-k)+1
y= 1/2(5(m-k)+1)

inserting some positive values of k,m we can find y and thereby y^2/5.

So, C is the answer.

GMAT Club team member

Joined: 02 Sep 2009

Posts: 11634

Followers: 1802

Kudos [?]: 9611 [11] , given: 829

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

26 Apr 2012, 22:11

This post received
KUDOS

If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?

(1) When x-y is divided by 5, the remainder is 1 --> x-y=5q+1, so x-y can be 1, 6, 11, ... Now, x=2 and y=1 (x-y=1) then x^2+y^2=5 and thus the remainder is 0, but if x=3 and y=2 (x-y=1) then x^2+y^2=13 and thus the remainder is 3. Not sufficient.

(2) When x+y is divided by 5, the remainder is 2 --> x+y=5p+2, so x+y can be 2, 7, 12, ... Now, x=1 and y=1 (x+y=2) then x^2+y^2=2 and thus the remainder is 2, but if x=5 and y=2 (x+y=7) then x^2+y^2=29 and thus the remainder is 4. Not sufficient.

(1)+(2) Square both expressions: x^2-2xy+y^2=25q^2+10q+1 and x^2+2xy+y^2=25p^2+20p+4 --> add them up: 2(x^2+y^2)=5(5q^2+2q+5p^2+4p+1) --> so 2(x^2+y^2) is divisible by 5 (remainder 0), which means that so is x^2+y^2. Sufficient.

Answer: C.

Hope it's clear.
_________________

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

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. NEW!!!

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. NEW!!!

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

Find out what's new at GMAT Club - latest features and updates

Intern

Joined: 28 Dec 2011

Posts: 10

Concentration: Economics, Strategy

Schools: Booth '15 (A), LBS '14 (A)

GMAT 1: 760 Q49 V46

GPA: 4

WE: Business Development (Investment Banking)

Followers: 0

Kudos [?]: 7 [0], given: 1

Re: if x and y are integers [#permalink]

16 May 2012, 11:59

raingary wrote:

1) x-y= 5k+1
tells us only about the difference between x and y , not about the individual numbers. hence not sufficient.
2) x+y=5m+2
tells us only about the addition of x and y , not about the individual numbers. hence not sufficient.
1&2) adding both the conditions:
2x= 5k+5m+3
2x= 5(k+m)+3
x= 1/2(5(k+m)+3)
inserting some positive values of k,m we can find x and thereby x^2/5.

for y subtract both the conditions:
2y= 5m+2 - 5k-1
2y= 5(m-k)+1
y= 1/2(5(m-k)+1)

inserting some positive values of k,m we can find y and thereby y^2/5.

So, C is the answer.

You've arrived at the correct answer, but your reasoning should be more on the grounds of reasoning why x-y leaving a remainder of 1 cannot provide sufficient information to ascertain if x^2+y^2 is divisible by 5.

Quick way to arrive to this in the exam is the following:

You know that 2(x^2+y^2)= (x-y)^2 + (x+y)^2. You've understood that x-y = 5q + 1 and that x+y = 5p+2. So, now we have:

2(x^2+y^2) = (5q+1)^2 + (5p+2)^2.

You know that expression is going to be divisible by 5....except for the units part. If the units part is divisible by 5, you're done. That's clearly the case just by looking at the expansion, as you will get 4+1.

So, as Bunuel put it, if twice the number is divisible by 5 (think 20), then the initial number (think 10) is also divisible by 5, and you're done.

GMAT Club team member

Joined: 02 Sep 2009

Posts: 11634

Followers: 1802

Kudos [?]: 9611 [0], given: 829

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

16 May 2012, 12:04

mila84 wrote:

Hello Bunuel

Do you know any similar questions, to practice?

Thank you

DS questions on remainders: search.php?search_id=tag&tag_id=198
PS questions on remainders: search.php?search_id=tag&tag_id=199

Hope it helps.
_________________

Senior Manager

Joined: 30 Jun 2011

Posts: 272

Followers: 0

Kudos [?]: 10 [0], given: 20

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

22 May 2012, 22:41

Bunuel wrote:

so 2(x^2+y^2) is divisible by 5 (remainder 0), which means that so is x^2+y^2. Sufficient.
Hope it's clear.

I didnt get above quote 2*5/2 is divisble by 5 bt 5/2 is not

GMAT Club team member

Joined: 02 Sep 2009

Posts: 11634

Followers: 1802

Kudos [?]: 9611 [0], given: 829

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

22 May 2012, 23:36

vikram4689 wrote:

Bunuel wrote:

so 2(x^2+y^2) is divisible by 5 (remainder 0), which means that so is x^2+y^2. Sufficient.
Hope it's clear.

I didnt get above quote 2*5/2 is divisble by 5 bt 5/2 is not

Correct example would be: 2*5 is divisible by 5 and so is 5.

2(x^2+y^2) to be divisible by 5, x^2+y^2 must be divisible by 5.

Hope it's clear.
_________________

Senior Manager

Joined: 30 Jun 2011

Posts: 272

Followers: 0

Kudos [?]: 10 [0], given: 20

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

22 May 2012, 23:50

Bunuel wrote:

vikram4689 wrote:

Bunuel wrote:

so 2(x^2+y^2) is divisible by 5 (remainder 0), which means that so is x^2+y^2. Sufficient.
Hope it's clear.

I didnt get above quote 2*5/2 is divisble by 5 bt 5/2 is not

Correct example would be: 2*5 is divisible by 5 and so is 5.

2(x^2+y^2) to be divisible by 5, x^2+y^2 must be divisible by 5.

Hope it's clear.

Both are correct examples but the one i gave does not follow what you mentioned

You said since 2*P is divisible by 5, therefore P is divisible by 5... This is not correct P=5/2

GMAT Club team member

Joined: 02 Sep 2009

Posts: 11634

Followers: 1802

Kudos [?]: 9611 [0], given: 829

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

22 May 2012, 23:55

vikram4689 wrote:

Both are correct examples but the one i gave does not follow what you mentioned

You said since 2*P is divisible by 5, therefore P is divisible by 5... This is not correct P=5/2

Not so. Your example is NOT correct.

I'm saying that if p is an integer and 2p is divisible by 5 then p must be divisible by 5. Now, check your example, is it correct?

It should be: 2*5 is divisible by 5 (p=5) so 5 (p) is divisible by 5.

Hope it's clear.
_________________

Intern

Joined: 19 Aug 2012

Posts: 1

Followers: 0

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

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

28 Aug 2012, 14:25

Question to Bunuel:

Please tell me, how in 2 minutes, looking at the question for the first time in my life, I can decifer that I shoud firstly square both equations, then ADD them and only then I will see that it can be divided by 5 too. where is the key to start such calculation? there is no time to do different approaches.

GMAT Club team member

Joined: 02 Sep 2009

Posts: 11634

Followers: 1802

Kudos [?]: 9611 [0], given: 829

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

29 Aug 2012, 01:17

rockfeld wrote:

We know that x-y=5q+1 and x+y=5p+2 and we need to find the remainder when x^2 + y^2 (x squared plus y squared ) is divided by 5, so we need to get some expression, from these two, where x and y are squared and add up. Squaring and adding seems to be the best way to proceed.

Hope it's clear.
_________________

Intern

Joined: 28 Aug 2012

Posts: 47

Location: Austria

GMAT 1: 770 Q51 V42

Followers: 3

Kudos [?]: 24 [0], given: 3

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

31 Aug 2012, 04:17

My approach, combining both statements:

(5n+r)^2=25n^2+10nr+r^2
25n^2 is divisible by 5, 10nr too. So we only have to square the remainder to find the remainder of the square of n. The same is true for adding and subtracting numbers, which should be clear.
Remainder of x is 1 greater than remainder of y (because of statement 1).
--> Options for remainders of x and y: (1,0), (2,1), (3,2), (4,3), (5,4)-->(0,4)
Statement 2 tells us to add x+y, so remainders of our options are: 1+0=1, 2+1=3, 3+2=5+0, 4+3=5+2, 0+4=4
Statement 2 says that the remainder is 2, when adding. Only (4,3) accomplishes this.
So 4^2+3^2=16+9=25
The remainder will always be 0 and both statements combined are therefore sufficient.

Director

Joined: 22 Mar 2011

Posts: 608

WE: Science (Education)

Followers: 43

Kudos [?]: 267 [0], given: 43

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

03 Sep 2012, 14:56

kt750 wrote:

If x and y are integer, what is the remainder when x^2 + y^2 is divided 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 2

Neither (1) nor (2) alone is sufficient.

(1) and (2) together:

From (1), x-y=5a+1 for some integer a, and from (2), x+y = 5b+2, for some integer b.
(x-y)(x+y) = x^2-y^2=(5a+1)(5b+2)=M5+2 (multiple of 5 plus 2).
Integers can be of the form M5, M5+1, M5-1(=M5+4), M5+2 or M5-2(=M5+3).
Therefore, square of an integer can be of the form M5, M5+1 or M5-1(=M5+4).

Since x^2-y^2 is M5+2, necessarily x^2 must be M5+1 and y^2 must be M5-1.
Then the sum x^2+y^2 is M5 (multiple of 5).
Sufficient.

Answer C
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Veritas Prep GMAT Instructor

Joined: 16 Oct 2010

Posts: 3114

Location: Pune, India

Followers: 573

Kudos [?]: 2020 [2] , given: 92

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

03 Sep 2012, 22:13

This post received
KUDOS

kt750 wrote:

If x and y are integer, what is the remainder when x^2 + y^2 is divided 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 2

The question can be done quickly if you realize that you need to get x^2 + y^2 using the terms x-y and/or x+y. Either term alone is not sufficient to represent x^2 + y^2 but using both,

2(x^2 + y^2) = (x + y)^2 + (x - y)^2
2(x^2 + y^2) = (5a + 1)^2 + (5b + 2)^2

Every term in (5a + 1)^2 and (5b + 2)^2 will be divisible by 5 except last ones i.e. 1^2 and 2^2.
1+4 = 5 so the remainder when 2(x^2 + y^2) is divided by 5 is 0. So x^2 + y^2 must be a multiple of 5 and the remainder when x^2 + y^2 is divided by 5 must be 0.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Manager

Joined: 10 Sep 2012

Posts: 137

Followers: 2

Kudos [?]: 14 [0], given: 17

If X and Y are integers, what is the remainder when X^2 + Y^ [#permalink]

23 Nov 2012, 16:31

If X and Y are integers, what is the remainder when X^2 + Y^2 is divided 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 2

GMAT Club team member

Joined: 02 Sep 2009

Posts: 11634

Followers: 1802

Kudos [?]: 9611 [0], given: 829

Re: If X and Y are integers, what is the remainder when X^2 + Y^ [#permalink]

24 Nov 2012, 05:24

anon1 wrote:

If X and Y are integers, what is the remainder when X^2 + Y^2 is divided 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 2

Merging similar topics. Please refer to the solutions above.
_________________

Intern

Joined: 19 Nov 2012

Posts: 1

Schools: HBS '15

Followers: 0

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

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

27 Jan 2013, 21:53

Bunuel wrote:

vikram4689 wrote:

Both are correct examples but the one i gave does not follow what you mentioned

You said since 2*P is divisible by 5, therefore P is divisible by 5... This is not correct P=5/2

I believe vikram4689 is correct. If we know that 2P is divisible by 5, then we only know that P is divisible by 2.5.

Let,
P=x^2+y^2 and
N=5a^2+2a+5b^2+4b+1

Therefore, we have already proven:

2P=5N
P=(5/2)N

The only way for P to be divisible by 5 is if N is even (contains a factor of 2). Unfortunately, we can't factor out a 2 from N. Therefore, whether or not N is even depends on the values of A and B.

By looking at the equation of N, we can see that it will be even only if A and B are not both positive or both negative.

It works out that the equations (x-y)=5a+1 and (x+y)=5b+2 can both be satisfied only when A and B are not both positive or both negative. This is why Bunuel's incorrect assumption still works in this case.

There is no way I could have figured all of this out in 2 or 3 minutes though.

Veritas Prep GMAT Instructor

Joined: 16 Oct 2010

Posts: 3114

Location: Pune, India

Followers: 573

Kudos [?]: 2020 [0], given: 92

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

27 Jan 2013, 22:32

matt8 wrote:

Bunuel wrote:

vikram4689 wrote:

Both are correct examples but the one i gave does not follow what you mentioned

You said since 2*P is divisible by 5, therefore P is divisible by 5... This is not correct P=5/2

Think again - It's not 'Bunuel's incorrect assumption'. It is given that x and y are integers so, given that P=x^2+y^2, P will be a positive integer.
Now that we know that P is an integer, if 2P is divisible by 5, P must be divisible by 5.

Given that a and b are positive integers and that 'a' is divisible by 'b', a MUST have b as a factor. It's a basic mathematical fact. When I say, 'a' is divisible by 'b', I mean 'a' is a multiple of 'b' which means 'b' is a factor of 'a'. Try some examples to convince yourself.

This statement has far reaching implications.
Think of questions like: Is 2^{100} divisible by 3? I will say 'no' without a thought.
Is 2^{10}*3^{100}*5 divisible by 7? Again, absolutely no!

To reiterate, if 'a' is to be divisible by 'b', a must have b as a factor. (talking about positive integers)
_________________

Senior Manager

Joined: 10 Oct 2012

Posts: 287

Followers: 4

Kudos [?]: 94 [0], given: 20

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink]

28 Jan 2013, 22:39

2*(x^2+y^2) = (x+y)^2+(x-y)^2
now,
x+y = 5k+2
x-y = 5m+1

(x+y)^2=25k^2+4+20k = 5(5k^2+4k)+4=5c+4

Similarly,

(x-y)^2=25m^2+1+10m=5(5m^2+2m)+1 = 5p+1

Hence, adding together the above 2 equations , we can get the remainder as;

5c+5p+5 = 5(p+c)+5=5(t+1)+0 = Multiple of 5. Hence the remainder is zero.

Hence answer = C.

Re: If x and y are integer, what is the remainder when x^2 + y^2 [#permalink] 28 Jan 2013, 22:39

		Similar topics	Author	Replies	Last post
Similar Topics:		What is the remainder when x^2+y^2 is divided by 5, where x	getzgetzu	3	06 May 2006, 05:31
		What is the value of x^2 - y^2 ?	redpearl	2	10 May 2012, 00:07
	1	What is the value of x^2 - 4y^2?	derekgmat	2	08 Nov 2012, 03:10
	2	What is the value of x^2+y^2 ?	maxLRok	8	23 Nov 2012, 07:29
	2	What is the remainder when x^2 - y^2 is divided by 3?	manishuol	2	29 Apr 2013, 12:05

If x and y are integer, what is the remainder when x^2 + y^2

Main navigation

GMAT Resources

Partners

MBA Resources

GMAT ® is a registered trademark of the Graduate Management Admission Council ® (GMAC ®). GMAT Club's website has not been reviewed or endorsed by GMAC.