Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 15 Jul 2019, 23:21 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If x and y are integer, what is the remainder when x^2 + y^2

Author Message
TAGS:

### Hide Tags

Director  Joined: 14 Dec 2012
Posts: 739
Location: India
Concentration: General Management, Operations
GMAT 1: 700 Q50 V34 GPA: 3.6
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

2
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

hi,

let a and b and c are 3 arbitray integers.

and if ==>a/c==>remainder x
and ===>b/c==>remainder is y

then
==>remainder of (a*b)/c=remainder of a*remainder of b
==>remainder of (a+b)/c=remainder of a+remainder of b
==>remainder of (a-b)/c=remainder of a-remainder of b

so using this clearly we can say that we need both the statement to solve this.
hence C
_________________
When you want to succeed as bad as you want to breathe ...then you will be successfull....

GIVE VALUE TO OFFICIAL QUESTIONS...

learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmat-analytical-writing-assessment
Intern  Joined: 26 Mar 2014
Posts: 4
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

from HTale's post on 17 May 2012: "You know that 2(x^2+y^2)= (x-y)^2 + (x+y)^2."

Math Expert V
Joined: 02 Sep 2009
Posts: 56234
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

1
meenanke wrote:
from HTale's post on 17 May 2012: "You know that 2(x^2+y^2)= (x-y)^2 + (x+y)^2."

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

Hope it helps.
_________________
Intern  Joined: 08 Oct 2015
Posts: 7
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

Bunuel 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 --> $$x-y=5q+1$$, 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.

Hope it's clear.

Question for Bunuel. Theoretically, could you have used only q or p to represent the qoutient in this example. Wouldn't they represent the same whole factor of times that 5 can go into each statement?
Retired Moderator Joined: 29 Oct 2013
Posts: 257
Concentration: Finance
GPA: 3.7
WE: Corporate Finance (Retail Banking)
If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

IMO this question is a walk in the park if we know the following rules:

You can always ignore the original number for remainder calculation. If you know a/n gives remainder p and b/n gives remainder of q then-
i) remainder of (a+b)/n = remainder of (p+q)/n
ii) remainder of (a-b)/n = remainder of (p-q)/n
iii) remainder of (a^2+b^2)/n = remainder of (p^2+q^2)/n
iv) remainder of (a^2-b^2)/n = remainder of (p^2-q^2)/n
v) remainder of (a*b)/n = remainder of (p*q)/n
vi) remainder of (a/b)/n = remainder of (p/q)/n
.
.
.
etc

But Im not quite sure about the accuracy of these rules.

So Moderators and Math Experts:

How do the above rules look neat or preposterous ?;) Any corrections, additions, exceptions? Thanks
_________________

My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876
CEO  S
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

1
NoHalfMeasures wrote:
IMO this question is a walk in the park if we know the following rules:

You can always ignore the original number for remainder calculation. If you know a/n gives remainder p and b/n gives remainder of q then-
i) remainder of (a+b)/n = remainder of (p+q)/n
ii) remainder of (a-b)/n = remainder of (p-q)/n
iii) remainder of (a^2+b^2)/n = remainder of (p^2+q^2)/n
iv) remainder of (a^2-b^2)/n = remainder of (p^2-q^2)/n
v) remainder of (a*b)/n = remainder of (p*q)/n
vi) remainder of (a/b)/n = remainder of (p/q)/n
.
.
.
etc

But Im not quite sure about the accuracy of these rules.

So Moderators and Math Experts:

How do the above rules look neat or preposterous ?;) Any corrections, additions, exceptions? Thanks

The best way would be to check them yourself by following the method shown below for (1).

You are given the following ,

a=nA+p and
b=nB+q

Thus, (a+b) = n(A+B) + p+q , or in other words, you will get a remainder of p+q when you divide a+b by n.

You can come up with similar relations based on method above. Additionally, you should not be remembering these relations and should be applying them as and when needed from first principles. Learning these obscure and not that common relations will only end up confusing you.
Intern  Joined: 25 Apr 2011
Posts: 5
Location: Pakistan
WE: Other (Commercial Banking)
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

2
Hi, this is my first post on this forum. I was little confused with the way I solved the problem and I don't know where did I go wrong!!

Statement 1 & 2 are insufficient individually..ok

When I combined the two statements, I multiplied the dividends and the remainders:

Dividends: (x-y) * (x+y) = x^2 - y^2
Remainders: 1 * 2 = 2

As we know that the divisor in this case is the same (i.e. 5) I assume we can multiply the different dividends and the remainders. In this way we can have the remainder (which is product of the two individual remainders) when x^2 - y^2 is divided by 5.

Once again: When (x^2 - y^2) / 5 the remainder is 2...

We can rewrite this as 5q + 2 = x^2 - y^2

Now I plugged in different values for q like 1,2,3,4

When q = 1, 5(1) + 2 = 7 -----> x^2 - y^2 = 7
When q = 2, 5(2) + 2 = 12 -----> x^2 - y^2 = 12
When q = 3, 5(3) + 2 = 17 -----> x^2 - y^2 = 17
When q = 4, 5(4) + 2 = 22 -----> x^2 - y^2 = 22

Now we can assume different values of x and y so that our total equals 7,12,17 and 22.

For the first case ( when x^2 - y^2 = 7) we can assume x^2=10 and y^2= 3 so when x^2 - y^2 = 7, then x^2 + y^2 = 10. In this case remainder will be 0 when x^2 + y^2 is divided by 5

For the second case ( when x^2 - y^2 = 12) we can assume x^2=15 and y^2= 3 so when x^2 - y^2 = 12, then x^2 + y^2 = 18. In this case remainder will be 3 when x^2 + y^2 is divided by 5

For the third case ( when x^2 - y^2 = 17) we can assume x^2=21 and y^2= 4 so when x^2 - y^2 = 17, then x^2 + y^2 = 25. In this case remainder will be 0 when x^2 + y^2 is divided by 5

For the fourth case ( when x^2 - y^2 = 22) we can assume x^2=25 and y^2= 3 so when x^2 - y^2 = 22, then x^2 + y^2 = 28. In this case remainder will be 3 when divided by 5

So we are not getting consistent remainders when x^2 + y^2 is divided by 5, aren't Statement 1 and 2 together also insufficient?
Manager  D
Joined: 17 May 2015
Posts: 248
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

1
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

Hi,

Clearly, statement 1 and 2 individually are not sufficient.

(1) When x-y is divided by 5, the remainder is 1

=> $$(x-y)^{2}$$ divided by 5, the remainder is 1. => $$x^{2} + y^{2} - 2xy = 5k + 1$$ ---(*)

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

=> $$(x+y)^{2}$$ divided by 5, the remainder is 4. => $$x^{2} + y^{2} + 2xy = 5m + 4$$ ---(**)

Add equation (*) and (**), we have following:

$$2*(x^{2} + y^{2}) = 5(k+m) + 5$$

Hence, when $$x^2 + y^2$$ is divided by 5, the remainder is 0.

Thanks.
Manager  B
Joined: 26 Mar 2017
Posts: 114
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

Bunuel 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 --> $$x-y=5q+1$$, 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.

Hope it's clear.

as usual a very good explanation

but how do I recognise when I have square the terms ??
_________________
I hate long and complicated explanations!
Intern  B
Joined: 25 Jul 2017
Posts: 16
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

Bunuel 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 --> $$x-y=5q+1$$, 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.

Hope it's clear.

Hi Bunuel,

When you squared each term, why did you make X=5 and Y=1 in the first expression and X=5 and Y=2 in the second expression? I don't understand what indicated that.
Math Expert V
Joined: 02 Sep 2009
Posts: 56234
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

bkastan wrote:
Bunuel 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 --> $$x-y=5q+1$$, 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.

Hope it's clear.

Hi Bunuel,

When you squared each term, why did you make X=5 and Y=1 in the first expression and X=5 and Y=2 in the second expression? I don't understand what indicated that.

Not following you... Which part are you talking about?
_________________
Intern  B
Joined: 25 Jul 2017
Posts: 16
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

Bunuel wrote:
bkastan wrote:
Bunuel 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 --> $$x-y=5q+1$$, 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.

Hope it's clear.

Hi Bunuel,

When you squared each term, why did you make X=5 and Y=1 in the first expression and X=5 and Y=2 in the second expression? I don't understand what indicated that.

Not following you... Which part are you talking about?

How did you get the following: (1)+(2) Square both expressions: x2−2xy+y2=25q2+10q+1x2−2xy+y2=25q2+10q+1 and x2+2xy+y2=25p2+20p+4x2+2xy+y2=25p2+20p+4 --> add them up: 2(x2+y2)=5(5q2+2q+5p2+4p+1)2(x2+y2)=5(5q2+2q+5p2+4p+1) --> so 2(x2+y2)2(x2+y2)

How did x^2 turn into 25Q^2?
Math Expert V
Joined: 02 Sep 2009
Posts: 56234
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

1
bkastan wrote:
How did you get the following: (1)+(2) Square both expressions: x2−2xy+y2=25q2+10q+1x2−2xy+y2=25q2+10q+1 and x2+2xy+y2=25p2+20p+4x2+2xy+y2=25p2+20p+4 --> add them up: 2(x2+y2)=5(5q2+2q+5p2+4p+1)2(x2+y2)=5(5q2+2q+5p2+4p+1) --> so 2(x2+y2)2(x2+y2)

How did x^2 turn into 25Q^2?

$$x-y=5q+1$$ --> $$(x-y)^2=(5q+1)^2=25q^2+10q+1$$

$$x+y=5p+2$$ --> $$(x+y)^2=(5p+2)^2=25p^2+20p+4$$
_________________
Intern  B
Joined: 25 Jul 2017
Posts: 16
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

Bunuel wrote:
bkastan wrote:
How did you get the following: (1)+(2) Square both expressions: x2−2xy+y2=25q2+10q+1x2−2xy+y2=25q2+10q+1 and x2+2xy+y2=25p2+20p+4x2+2xy+y2=25p2+20p+4 --> add them up: 2(x2+y2)=5(5q2+2q+5p2+4p+1)2(x2+y2)=5(5q2+2q+5p2+4p+1) --> so 2(x2+y2)2(x2+y2)

How did x^2 turn into 25Q^2?

$$x-y=5q+1$$ --> $$(x-y)^2=(5q+1)^2=25q^2+10q+1$$

$$x+y=5p+2$$ --> $$(x+y)^2=(5p+2)^2=25p^2+20p+4$$

Thank you so much. You're so smart!!
Current Student P
Joined: 18 Aug 2016
Posts: 617
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29 GMAT 2: 740 Q51 V38 Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

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

In such questions for me it is more easier to solve with actual numbers

by seeing the question and reading the options you can easily sense that option A, B, D are wrong (choice will most probably be in between C & E)

Identify numbers for which conditions support
(1) x,y could be 7,6 then remainder will be 0
x,y could be 8,7 then remainder will be 3..not sufficient

(2)x,y could be 8,4 then remainder will be 0
x,y could be 7,5 then remainder will be 4..not sufficient

However on combining we can see that when x,y is 9,8 it passes both the conditions and the remainder is 0
We can check as well
when x,y is 4,3 it passes both the conditions and the remainder is 0
Hence C
_________________
We must try to achieve the best within us

Thanks
Luckisnoexcuse
Manager  B
Joined: 02 Jul 2017
Posts: 67
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

x-y =5a+1
x+y =5b+2

statement 1 :
can't tell about remainder of .. x^2+y^2=(x-y)^2 +2xy

statement 2
can't tell about remainder of x^2+y^2=(x+y)^2 -2xy

as indicated in statement 1 and 2 ..need to get rid of xy lets us add staement 1 and 2

2x^2+2y^2=5( some expression) ....

so 2( x^2+y^2) will be divisible , hence answer C
Manager  G
Joined: 30 Mar 2017
Posts: 130
GMAT 1: 200 Q1 V1 Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

1
I can offer an alternate approach:

To find the remainder when x^2 + y^2 is divided by 5, we can focus on the units digits of x and y.

From Statement 1, we can deduce that (x-y) will have a units digit of 1 or 6. So, possible units digits of (x,y) are (2,1), (3,2), (8,2), etc. Not sufficient.
From Statement 2, we can deduce that (x+y) will have a units digit of 2 or 7. So, possible units digits of (x,y) are (1,1), (9,3), (5,2), etc. Not sufficient.

Taking both statements together, we have the following 4 possibilities:

Case 1
x-y=1
x+y=2
-->2x=3
-->x is not an integer; discard

Case 2
x-y=1
x+y=7
-->2x=8
-->x is an integer, so possibly we have the answer.

Case 3
x-y=6
x+y=2
-->2x=8
-->same as Case 2

Case 4
x-y=6
x+y=7
-->2x=13
-->x is not an integer; discard

So units digit of (x,y) is (4,3) --> units digit of (x^2,y^2) is (6,9) --> units digit of x^2 + y^2 is 5. Thus remainder is 0 when divided by 5.

Non-Human User Joined: 09 Sep 2013
Posts: 11649
Re: If x and y are integer, what is the remainder when x^2 + y^2  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: If x and y are integer, what is the remainder when x^2 + y^2   [#permalink] 10 Apr 2019, 16:41

Go to page   Previous    1   2   [ 38 posts ]

Display posts from previous: Sort by

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