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 Your Progress

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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

06 Jul 2013, 07:52

2

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

(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...

GMAT RCs VOCABULARY LIST: http://gmatclub.com/forum/vocabulary-list-for-gmat-reading-comprehension-155228.html learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmat-analytical-writing-assessment : http://www.youtube.com/watch?v=APt9ITygGss

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

Show Tags

28 Mar 2015, 10:04

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]

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 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.

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?

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 (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
_________________

Please contact me for super inexpensive quality private tutoring

My journey V46 and 750 -> http://gmatclub.com/forum/my-journey-to-46-on-verbal-750overall-171722.html#p1367876

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.
_________________

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

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?

gmatclubot

Re: If x and y are integer, what is the remainder when x^2 + y^2
[#permalink]
23 Aug 2016, 05:19

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Term 1 has begun. If you're confused, wondering what my post on the last 2 official weeks was, that was pre-term. What that means is that the school...