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



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



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



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



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



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Re: If x and y are integer, what is the remainder when x^2 + y^2
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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 ??
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Re: If x and y are integer, what is the remainder when x^2 + y^2
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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.



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



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



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



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




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