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, the remainder is 2
OA C
x^2 + y^2 divided by 5
This topic has expert replies
- Shalabh's Quants
- Master | Next Rank: 500 Posts
- Posts: 134
- Joined: Fri Apr 06, 2012 3:11 am
- Thanked: 35 times
- Followed by:5 members
As is clear that from Statement 1 and 2 that by alone, we cannot infer the result.massi2884 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, the remainder is 2
OA C
So Lets combine.
To make the calculations easier, assume x - y = 1 & x + y = 5k + 2. Where k is a natural no.
This yields x = (5k+3)/2 & y = (5k+1)/2.
So, x^2 + y^2 = [(5k+3)/2]^2 + [(5k+1)/2]^2 = 1/4.(50 k^2 + 40 K + 10); This expression is multiple of 5, hence remainder will be 0. Answer C.
Shalabh Jain,
e-GMAT Instructor
e-GMAT Instructor
- neelgandham
- Community Manager
- Posts: 1060
- Joined: Fri May 13, 2011 6:46 am
- Location: Utrecht, The Netherlands
- Thanked: 318 times
- Followed by:52 members
If x and y are integers, what is the remainder when x^2 + y^2 is divided by 5?
If x = 4 and y = 3 then x^2 + y^2 = 16 + 9 = 25 and the remainder when x^2 + y^2 is divided by 5 is 0
If x = 4 and y = 3 then x^2 + y^2 = 16 + 9 = 25 and the remainder when x^2 + y^2 is divided by 5 is 0
2) When x+y is divided by, the remainder is 2. Implies that x + y is of the form 5l + 2, where l is a non negative integer.
So, x + y = 5l + 2and x - y = 5k + 1
Squaring and adding, we get
x^2 + y^2 + 2xy + x^2 + y^2 - 2xy = 25*l^2 + 4 + 20*l + 25*k^2 + 1 + 10k
2*(x^2 + y^2) = 25*l^2 + 20*l + 25*k^2 + 10k + 5
2*(x^2 + y^2) = 5(5*l^2 + 4*l + 5*k^2 + 2k + 1)
2*(x^2 + y^2) = 5* An integer
So, we can say that x^2 + y^2 is a multiple of 5 and that if x^2 + y^2 is divided by 5, it leaves a remainder of 0
Hence C
If x = 6 and y = 5 then x^2 + y^2 = 36 + 25 and the remainder when x^2 + y^2 is divided by 5 is 11) When x-y is divided by 5, the remainder is 1
If x = 4 and y = 3 then x^2 + y^2 = 16 + 9 = 25 and the remainder when x^2 + y^2 is divided by 5 is 0
If x = 7 and y = 5 then x^2 + y^2 = 49 + 25 and the remainder when x^2 + y^2 is divided by 5 is 32) When x+y is divided by, the remainder is 2
If x = 4 and y = 3 then x^2 + y^2 = 16 + 9 = 25 and the remainder when x^2 + y^2 is divided by 5 is 0
1) When x-y is divided by 5, the remainder is 1. Implies that x - y is of the form 5k + 1, where k is a non negative integer.From 1 + 2
2) When x+y is divided by, the remainder is 2. Implies that x + y is of the form 5l + 2, where l is a non negative integer.
So, x + y = 5l + 2and x - y = 5k + 1
Squaring and adding, we get
x^2 + y^2 + 2xy + x^2 + y^2 - 2xy = 25*l^2 + 4 + 20*l + 25*k^2 + 1 + 10k
2*(x^2 + y^2) = 25*l^2 + 20*l + 25*k^2 + 10k + 5
2*(x^2 + y^2) = 5(5*l^2 + 4*l + 5*k^2 + 2k + 1)
2*(x^2 + y^2) = 5* An integer
So, we can say that x^2 + y^2 is a multiple of 5 and that if x^2 + y^2 is divided by 5, it leaves a remainder of 0
Hence C
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
- charu_mahajan
- Master | Next Rank: 500 Posts
- Posts: 111
- Joined: Tue Jan 31, 2012 1:06 pm
- Thanked: 15 times
- Followed by:8 members
If (x-y)/5 -> gives a reminder 1. That means that (x-y) should be some number like 6, 11, 16, 21...etc.1) When x-y is divided by 5, the remainder is 1
6/5 -> Remainder 1
11/5 -> Remainder 1
16/5 -> Remainder 1
If (x+y)/5 -> gives a reminder 2. That means that (x+y) should be some number like 7, 12, 17, 22...etc.2) When x+y is divided by 5, the remainder is 2
7/5 -> Remainder 2
12/5 -> Remainder 2
17/5 -> Remainder 2
Plugging in the valuesFrom 1 + 2
1) if x-y = 6 (Can be 6,11,16,21...)
x+y = 12(Can be 7,12,17,22...)
Solving for x
2x = 18 x=9
if x=9, y = 3
(x^2 + y^2)/5 = (9^2 + 3^2)/5 = (81+9)/5 = 90/5 -> Remainder =0
2) Cross check with another set of values
if x-y = 11 (Can be 6,11,16, 21...)
x+y = 17 (Can be 7,12,17,22...)
Solving for x
2x = 28 x=14 y=3
(14^2 + 3^2)/5 = (196 + 9)/5 = 205/5 -> Remainder =0
Hence Together sufficient -> C[/quote]
- vikram4689
- Legendary Member
- Posts: 1325
- Joined: Sun Nov 01, 2009 6:24 am
- Thanked: 105 times
- Followed by:14 members
..
Last edited by vikram4689 on Tue May 22, 2012 11:05 pm, edited 1 time in total.
Premise: If you like my post
Conclusion : Press the Thanks Button
Conclusion : Press the Thanks Button
GMAT/MBA Expert
- Anurag@Gurome
- GMAT Instructor
- Posts: 3835
- Joined: Fri Apr 02, 2010 10:00 pm
- Location: Milpitas, CA
- Thanked: 1854 times
- Followed by:523 members
- GMAT Score:770
(1) When x - y is divided by 5, the remainder is 1.massi2884 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, the remainder is 2
OA C
x - y = 5a + 1, so x - y can be 1, 6, 11, ...
If x = 2, y = 1, x - y = 1, then x² + y² = 5. So, remainder = 0.
If x = 3, y = 2, x - y = 1, then x² + y² = 13. So, remainder = 3.
No definite answer; NOT sufficient.
(2) When x + y is divided by, the remainder is 2.
x + y = 5b + 2, so x + y can be 2, 7, 12, ...
If x = 1, y = 1, x + y = 2, then x² + y² = 2. So, remainder = 2.
If x = 5, y = 2, x + y = 7, then x² + y² = 29. So, remainder = 4.
No definite answer; NOT sufficient.
Combining (1) and (2), x - y = 5a + 1 and x + y = 5b + 2
(x - y)² = (5a + 1)² or x² - 2xy + y² = 25a² + 10a + 1
(x + y)² = (5b + 2)² or x² + 2xy + y² = 25b² + 20b + 4
Adding the 2 equations, we get
2(x² + y²) = 5(5a² + 5b² + 2a + 4b + 1), which clearly implies that 2(x² + y²) is divisible by 5 with remainder = 0 and so x² + y² is also divisible by 5 with remainder = 0; SUFFICIENT.
The correct answer is C.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)
Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)
Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
- ronnie1985
- Legendary Member
- Posts: 626
- Joined: Fri Dec 23, 2011 2:50 am
- Location: Ahmedabad
- Thanked: 31 times
- Followed by:10 members
-
- Senior | Next Rank: 100 Posts
- Posts: 79
- Joined: Tue Dec 15, 2015 3:56 am
- Followed by:1 members
- GMAT Score:750
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, Experts:
How do the above rules look neat or preposterous ?;) Any corrections, additions, exceptions? Thanks
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, Experts:
How do the above rules look neat or preposterous ?;) Any corrections, additions, exceptions? Thanks
-
- Senior | Next Rank: 100 Posts
- Posts: 79
- Joined: Tue Dec 15, 2015 3:56 am
- Followed by:1 members
- GMAT Score:750
Can some expert comment on this, please? Thanks
Nina1987 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, Experts:
How do the above rules look neat or preposterous ?;) Any corrections, additions, exceptions? Thanks
GMAT/MBA Expert
- [email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
HI Nina1987,
One of the great aspects of the GMAT is that you can approach most questions in a variety of different ways. To that end, 'your way' of doing things is fine as long as it helps you to hit your score goals.
1) How have you been scoring on your CATs (including the Quant and Verbal Scaled Scores)?
2) What is your goal score?
3) How long have you been studying?
4) When are you planning to take the GMAT?
GMAT assassins aren't born, they're made,
Rich
One of the great aspects of the GMAT is that you can approach most questions in a variety of different ways. To that end, 'your way' of doing things is fine as long as it helps you to hit your score goals.
1) How have you been scoring on your CATs (including the Quant and Verbal Scaled Scores)?
2) What is your goal score?
3) How long have you been studying?
4) When are you planning to take the GMAT?
GMAT assassins aren't born, they're made,
Rich
- Max@Math Revolution
- Elite Legendary Member
- Posts: 3991
- Joined: Fri Jul 24, 2015 2:28 am
- Location: Las Vegas, USA
- Thanked: 19 times
- Followed by:37 members
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
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
In the original condition, there are 2 variables(x,y), which should match with the number of equations. So you need 2 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1) & 2), it becomes x-y=1,6,11,..... x+y=2,7,12... and you get x=4, y=3 from it. So, from x^2+y^2=4^2+3^2=25, the remainder of 25 divided by 5 is 0, which is unique and sufficient. Therefore, the answer is C.
-> For cases where we need 2 more equations, such as original conditions with "2 variables", or "3 variables and 1 equation", or "4 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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
In the original condition, there are 2 variables(x,y), which should match with the number of equations. So you need 2 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer. In 1) & 2), it becomes x-y=1,6,11,..... x+y=2,7,12... and you get x=4, y=3 from it. So, from x^2+y^2=4^2+3^2=25, the remainder of 25 divided by 5 is 0, which is unique and sufficient. Therefore, the answer is C.
-> For cases where we need 2 more equations, such as original conditions with "2 variables", or "3 variables and 1 equation", or "4 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
Math Revolution
The World's Most "Complete" GMAT Math Course!
Score an excellent Q49-51 just like 70% of our students.
[Free] Full on-demand course (7 days) - 100 hours of video lessons, 490 lesson topics, and 2,000 questions.
[Course] Starting $79 for on-demand and $60 for tutoring per hour and $390 only for Live Online.
Email to : [email protected]
I substituting number is a time consuming and lengthy process. Atleast for me My way of solving this
(X+Y) =5m+1
X-Y = 5m+2
We need x ^2 +Y^2
and when we square it all the variables will contain 5m except the last constant .
so without expannding it we will knw the value .
i.e
1+4 =5 remeinder of 5 is zero
option C
(X+Y) =5m+1
X-Y = 5m+2
We need x ^2 +Y^2
and when we square it all the variables will contain 5m except the last constant .
so without expannding it we will knw the value .
i.e
1+4 =5 remeinder of 5 is zero
option C