Jul 16 03:00 PM PDT  04:00 PM PDT Join a free live webinar and find out which skills will get you to the top, and what you can do to develop them. Save your spot today! Tuesday, July 16th at 3 pm PST Jul 16 08:00 PM EDT  09:00 PM EDT Strategies and techniques for approaching featured GMAT topics. Tuesday, July 16th at 8 pm EDT Jul 15 10:00 PM PDT  11:00 PM PDT Get the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Jul 15 10:00 PM PDT  11:00 PM PDT Take 20% off the plan of your choice, now through midnight on Monday, July 15th Jul 19 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes
Author 
Message 
TAGS:

Hide Tags

Director
Joined: 14 Dec 2012
Posts: 739
Location: India
Concentration: General Management, Operations
GPA: 3.6

Re: If x and y are integer, what is the remainder when x^2 + y^2
[#permalink]
Show Tags
06 Jul 2013, 07:52
If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5? (1) When xy 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 (ab)/c=remainder of aremainder 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...
GMAT RCs VOCABULARY LIST: http://gmatclub.com/forum/vocabularylistforgmatreadingcomprehension155228.html learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmatanalyticalwritingassessment : http://www.youtube.com/watch?v=APt9ITygGss



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
26 Mar 2014, 22:26
from HTale's post on 17 May 2012: "You know that 2(x^2+y^2)= (xy)^2 + (x+y)^2."
Please help: how does 2(x^2+y^2) factor to (xy)^2 + (x+y)^2?



Math Expert
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
27 Mar 2014, 01:17
meenanke wrote: from HTale's post on 17 May 2012: "You know that 2(x^2+y^2)= (xy)^2 + (x+y)^2."
Please help: how does 2(x^2+y^2) factor to (xy)^2 + (x+y)^2? \((xy)^2 + (x+y)^2=(x^22xy+y^2)+(x^2+2xy+y^2)=2(x^2+y^2)\) 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
13 Nov 2015, 15:11
Bunuel wrote: If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?
(1) When xy is divided by 5, the remainder is 1 > \(xy=5q+1\), so \(xy\) can be 1, 6, 11, ... Now, \(x=2\) and \(y=1\) (\(xy=1\)) then \(x^2+y^2=5\) and thus the remainder is 0, but if \(x=3\) and \(y=2\) (\(xy=1\)) 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: \(x^22xy+y^2=25q^2+10q+1\) 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. 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
14 Dec 2015, 07:03
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 (ab)/n = remainder of (pq)/n iii) remainder of (a^2+b^2)/n = remainder of (p^2+q^2)/n iv) remainder of (a^2b^2)/n = remainder of (p^2q^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
_________________
Please contact me for super inexpensive quality private tutoring
My journey V46 and 750 > http://gmatclub.com/forum/myjourneyto46onverbal750overall171722.html#p1367876



CEO
Joined: 20 Mar 2014
Posts: 2620
Concentration: Finance, Strategy
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
16 Dec 2015, 12:46
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 (ab)/n = remainder of (pq)/n iii) remainder of (a^2+b^2)/n = remainder of (p^2+q^2)/n iv) remainder of (a^2b^2)/n = remainder of (p^2q^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
Concentration: Entrepreneurship, International Business
WE: Other (Commercial Banking)

Re: If x and y are integer, what is the remainder when x^2 + y^2
[#permalink]
Show Tags
23 Aug 2016, 05:19
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: (xy) * (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
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
23 Apr 2017, 01:10
kt750 wrote: If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?
(1) When xy 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 xy is divided by 5, the remainder is 1 => \((xy)^{2}\) 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. Sufficient. Answer(C) Thanks.



Manager
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
18 Jun 2017, 13:32
Bunuel wrote: If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?
(1) When xy is divided by 5, the remainder is 1 > \(xy=5q+1\), so \(xy\) can be 1, 6, 11, ... Now, \(x=2\) and \(y=1\) (\(xy=1\)) then \(x^2+y^2=5\) and thus the remainder is 0, but if \(x=3\) and \(y=2\) (\(xy=1\)) 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: \(x^22xy+y^2=25q^2+10q+1\) 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. as usual a very good explanation but how do I recognise when I have square the terms ??
_________________
I hate long and complicated explanations!



Intern
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
16 Nov 2017, 17:06
Bunuel wrote: If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?
(1) When xy is divided by 5, the remainder is 1 > \(xy=5q+1\), so \(xy\) can be 1, 6, 11, ... Now, \(x=2\) and \(y=1\) (\(xy=1\)) then \(x^2+y^2=5\) and thus the remainder is 0, but if \(x=3\) and \(y=2\) (\(xy=1\)) 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: \(x^22xy+y^2=25q^2+10q+1\) 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. 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
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
16 Nov 2017, 20:43
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 xy is divided by 5, the remainder is 1 > \(xy=5q+1\), so \(xy\) can be 1, 6, 11, ... Now, \(x=2\) and \(y=1\) (\(xy=1\)) then \(x^2+y^2=5\) and thus the remainder is 0, but if \(x=3\) and \(y=2\) (\(xy=1\)) 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: \(x^22xy+y^2=25q^2+10q+1\) 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. 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
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
16 Nov 2017, 20:45
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 xy is divided by 5, the remainder is 1 > \(xy=5q+1\), so \(xy\) can be 1, 6, 11, ... Now, \(x=2\) and \(y=1\) (\(xy=1\)) then \(x^2+y^2=5\) and thus the remainder is 0, but if \(x=3\) and \(y=2\) (\(xy=1\)) 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: \(x^22xy+y^2=25q^2+10q+1\) 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. 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
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
16 Nov 2017, 20:49
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? \(xy=5q+1\) > \((xy)^2=(5q+1)^2=25q^2+10q+1\) \(x+y=5p+2\) > \((x+y)^2=(5p+2)^2=25p^2+20p+4\)
_________________



Intern
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
16 Nov 2017, 21:55
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? \(xy=5q+1\) > \((xy)^2=(5q+1)^2=25q^2+10q+1\) \(x+y=5p+2\) > \((x+y)^2=(5p+2)^2=25p^2+20p+4\) Thank you so much. You're so smart!!



Current Student
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
17 Nov 2017, 00:13
kt750 wrote: If x and y are integer, what is the remainder when x^2 + y^2 is divided by 5?
(1) When xy 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
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
08 Dec 2017, 02:15
xy =5a+1 x+y =5b+2
statement 1 : can't tell about remainder of .. x^2+y^2=(xy)^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
Joined: 30 Mar 2017
Posts: 130

Re: If x and y are integer, what is the remainder when x^2 + y^2
[#permalink]
Show Tags
08 Apr 2018, 06:30
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 (xy) 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 xy=1 x+y=2 >2x=3 >x is not an integer; discard
Case 2 xy=1 x+y=7 >2x=8 >x is an integer, so possibly we have the answer.
Case 3 xy=6 x+y=2 >2x=8 >same as Case 2
Case 4 xy=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.
Answer: C



NonHuman 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
10 Apr 2019, 16:41
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 ]



