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: Remainder Problem, divisible by 5? [#permalink]

Show Tags

14 Nov 2009, 20:01

Hi,

I'm not sure if this is correct, but my answer is D. Both put together can be used to answer the question.

Explanation: Since 1 says x-y gives a remainder of 1 when divided by 5, this gives an option that x-y has to end in either 3 or 8. 2 says that x+y gives a remainder of 3 when divided by 5. Hence x+y has to end in either 1 or 6. If we try the possible combinations for x and y then it can only be when they end with 2 and 1 respectively. So if we square up numbers ending with 2 and 1 the unit digits will be 4 and 1 and hence sum to 5 which is divisible by 5.

Re: Remainder Problem, divisible by 5? [#permalink]

Show Tags

14 Nov 2009, 20:07

avinarvind wrote:

Hi,

I'm not sure if this is correct, but my answer is D. Both put together can be used to answer the question.

Explanation: Since 1 says x-y gives a remainder of 1 when divided by 5, this gives an option that x-y has to end in either 3 or 8. 2 says that x+y gives a remainder of 3 when divided by 5. Hence x+y has to end in either 1 or 6. If we try the possible combinations for x and y then it can only be when they end with 2 and 1 respectively. So if we square up numbers ending with 2 and 1 the unit digits will be 4 and 1 and hence sum to 5 which is divisible by 5.

Hence x^2+y^2 is divisible by 5.

Please correct me if I am wrong

I'm getting answer E

1. x-y/5 gives remainder 1

therefore x-y=1, 6, 11, 16, 21 insufficient because we don't know x or y

Re: Remainder Problem, divisible by 5? [#permalink]

Show Tags

14 Nov 2009, 22:55

I am not sure if I understood the above explanation.

But from what I see, the question is whether x^2+y^2 is divsible by 5.

So as per the explanation x^2 + y^2 is divisible by 5 since y=1 and x has to end in 7. The unit digit will always be 9+1=0 and hence divisible by 5. So answer is 'D'.

Re: Remainder Problem, divisible by 5? [#permalink]

Show Tags

15 Nov 2009, 05:10

Is x^2+y^2 divisible 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 3

Answer:

Statement 1: This tells us two things- 1) Either one of x and y is a multiple of 5 or 2) Both x and y are non -multiples of 5-( x-y=can result in multiple of 5 or non multiple of 5 ) Both x and y CANNOT be multiples of 5 -Not sufficient

Statement 2: This tells us the same thing as 1 1) Either one of x and y is a multiple of 5 or 2) Both x and y are non -multiples of 5( x+y=can result in multiple of 5 or non multiple of 5 ) Both x and y CANNOT be multiples of 5 -Not sufficient

Combining (1) and (2) Either one of x and y is a multiple of 5 or Both x and y are non -multiples of 5 this means either x2 or y2 is a multiple of 5 or both are non multiples of 5 x2+y2- this can result in a multiple of 5 or a non multiple - so Not sufficient

Re: Remainder Problem, divisible by 5? [#permalink]

Show Tags

15 Nov 2009, 08:38

6

This post received KUDOS

Expert's post

4

This post was BOOKMARKED

ctrlaltdel wrote:

Is x^2+y^2 divisible 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 3

First I must say this is not GMAT question. As every GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

But let's deal with question as it's written. x^2+y^2 to be divisible by 5 it should be an integer first and then should have the last digit 0 or 5:

(1) \(x-y=5n+1\) --> \((x-y)^2=25n^2+10n+1\) --> \(x^2+y^2=25n^2+10n+1+2xy\). Now 25n^2+10n is divisible by 5, but what about 1-2xy? Not sufficient.

(2) \(x+y=5m+3\) --> \((x+y)^2=25m^2+30m+9\) --> \(x^2+y^2=25m^2+30m+9-2xy\). The same here. Not sufficient.

(1)+(2) Add the equations: \(2(x^2+y^2)=25n^2+10n+25m^2+30m+10=25(n^2+m^2)+10(n+3m)+10\).

\(x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+3m)+5\). Well if \(n^2+m^2\) is even, \(x^2+y^2\) is an integer and it's divisible by 5. But what if \(n^2+m^2\) is odd, then \(x^2+y^2\) is not an integer at all, hence not divisible by 5.

Can we somehow conclude that \(n^2+m^2\) is even? If we are not told that x and y are integers then not.

Answer: E.

To check this consider following pairs of x an y: x=4.5, y=3.5 and x=7, y=1, \(x=4.5\), \(y=3.5\) \(x-y=4.5-3.5=1\) first statement holds true, 1 divided by 5 remainder 1. \(x+y=4.5+3.5=8\) second statement holds true, 8 divided by 5 remainder 3. \(x^2+y^2=4.5^2+3.5^2=32.5\) not divisible by 5.

\(x=7\), \(y=1\) \(x-y=7-1=6\) first statement holds true, 6 divided by 5 remainder 1. \(x+y=7+1=8\) second statement holds true, 8 divided by 5 remainder 3. \(x^2+y^2=7^2+1^2=50\) is divisible by 5.

Two different answers. Not sufficient. Answer: E.

Now let's see if we were told that x and y are positive integers, as GMAT does. Still would be quite hard 700+ question.

Last step: \(x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+3m)+5\). Can we now conclude that that \(n^2+m^2\) is even?

\(x-y=5n+1\) and \(x+y=5m+3\). Add them up: \(2x=5(n+m)+4\) --> \(x=\frac{5(n+m)}{2}+2\). As x is an integer n+m must be divisible by 2, hence either both are even or both are odd, in any case \(n^2+m^2\) is even and divisible by 2. Hence \(x^2+y^2=\frac{25(n^2+m^2)}{2}+5(n+15m)+5\) is an integer and divisible by 5.

x-y=5n+1 and x+y=5m+3. Add them up: 2x=5(n+m)+4 --> \(x=\frac{5(n+m)}{2}+2\). As x is an integer n+m must be divisible by 2, hence either both are even or both are odd, in any case n^2+m^2 is even and divisible by 2. Hence \(x^2+y^2=\frac{25(n^2+m^2)}{2} + 5(n+3m) + 5\) is an integer and divisible by 5.

Once we assume the question states that x and y are +ve integers, x^2 and y^2 have to be integers. In that case can we not skip all of the above, because the underlined rule applies for (n^2+m^2) in the aforementioned Last step ( i.e \(x^2+y^2=\frac{25(n^2+m^2)}{2} + 5(n+3m) + 5\)) too, right? I mean, As x^2+y^2 is an integer, \((n^2+m^2)\) must be to be divisible by 2, right?

Thank you

gmatclubot

Re: Is x^2+y^2 divisible by 5?
[#permalink]
10 Nov 2014, 01:27

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...