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:

If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5. (2) x – y = 3

For (1), the minimum value for p should 13(since 8+5/8 = 5). In this case, x can only be 2 (since 2x2 + 3x3 = 13). so x is not divisible by 4. The next possible value for p would be 29(since 8x3+5/8 = 5). Here, x again can only be 3 (since 2x2 + 5x5 = 29). so x is not divisible by 4.

So (1) is sufficient.

For (2), if x-y = 3, it be x should atleast be 4 since y is odd integer(4-1=3). In this case, x is divisible by 4. However, x can be 6(since 6-3=3), in that case its not divisible.

If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5. (2) x – y = 3

(1) If the remainder is 5 when p is divided by 8, x (or x^2) has to be an even integer since y is already said to be an odd integer. However its not sure whether x could be 2 or 4 or 6 or 8 or so on...

If x = 2, y could be any odd integer. Lets say y = 1, p = x^2 + y^2 = 5, which has 5 reminder when it is divided by 8. If x = 2 and y = 3, p = x^2 + y^2 = 13, which has 5 reminder when it is divided by 8. If x = 2, and y = 5, p = x^2 + y^2 = 29, which has 5 reminder when it is divided by 8.

But if x = 4, none of the values of y generates 5 reminder when p is divided by 8. For ex:

If x = 4, and y = 1, p = x^2 + y^2 = 17, which has 1 reminder when it is divided by 8. If x = 4, and y = 3, p = x^2 + y^2 = 25, which has 1 reminder when it is divided by 8. If x = 4, and y = 5, p = x^2 + y^2 = 41, which has 1 reminder when it is divided by 8. If x = 8, and y = 1, p has 1 reminder when it is divided by 8. If x = 8, and y = 3, p has 1 reminder when it is divided by 8. If x = 8, and y = 5, p has 1 reminder when it is divided by 8.

Any value for x i.e. divisible by 4 doesnot produce 5 reminder when p is divided by 8.

So it is sufficient to answer that x is not divisible by 4.

(2) If x = y+3, p = x^2 + y^2 = (y+3)^2 + y^2. Here y could be 1, or 3 or 5 or 7 or so on.... If y = 1, x = 4...........yes. If y = 3, x = 6...........no. If y = 5, x = 8........... yes.. If y = 7, x = 10.............no.................NSF...

(2) -> x is even and y is odd - square of even (x) is even and odd (y) is odd. sum of even + odd is always odd -> IT is not divisible by 4.

I will go for 'D'.

your statemet that "x is even" is not correct?

I think the statement that x is even is correct. From the statement 2, x-y=3 => x=y+3. 3 is odd and y is odd, then the sum of two odds is even => x is even

(2) -> x is even and y is odd - square of even (x) is even and odd (y) is odd. sum of even + odd is always odd -> IT is not divisible by 4.

I will go for 'D'.

your statemet that "x is even" is not correct?

I think the statement that x is even is correct. From the statement 2, x-y=3 => x=y+3. 3 is odd and y is odd, then the sum of two odds is even => x is even

agree X can be even ..but it is not sufficient.(2). Kudos to Gmat tiger for a brilliant explanation.

yes, statement 2 alone is not sufficient because the only thing we can deduct from it is that x is even but an integer can still be an even but not a multiple of 8. Statement 1 alone is sufficient: Assumption: x is a multiple of 4. If x is a multiple of 4, then x=4m=>x^2=16(m^2) p=16(m^2)+y^2. p/8=16(m^2)/8 + y^2/8 16(m^2) is divisible by 8. Therefore the remainder of p when divided by 8 will be equal to the remainer of y^2 when divided by 8. y is odd => y=2n+1=> y^2=(2n+1)^2=4n^2+4n+1=4n(n+1)+1 We have two subsequent numbers n and n+1. So one of them is even and 4n(n+1) is divisible by 8. y^2=8k+1. When a square of an odd number is divided by 8, the remainder will be 1. However, we are given that the remainder is 5 => the assumtion that x is a multiple of 4 is wrong. X cannot be a multiple of 4.

Last edited by LenaA on 26 Aug 2009, 10:04, edited 1 time in total.

All, I di agree with answer A However I am having issues understanding the option that x may be 2

To me, p=x^2+y^2 as long as y is odd and positive, its min value is 1 to me x can note be 0 as it is not +ve or -ve so p>=13 x^2 >= 12 so to me the min value of x is 4.

I am sure that I am missing something here. Can anyone help? Thx

All, I di agree with answer A However I am having issues understanding the option that x may be 2

To me, p=x^2+y^2 as long as y is odd and positive, its min value is 1 to me x can note be 0 as it is not +ve or -ve so p>=13 x^2 >= 12 so to me the min value of x is 4.

I am sure that I am missing something here. Can anyone help? Thx

No you are not missing. I made a mistake. I will edit my previous post. x cannot be 2 given the fact that x-y=3. The answer is still A though.Thanks for a correction.

Given p,x,y>0; y is odd - of the form (2k-1); p = \(x^2+y^2\) Is x divisible by 4? Stat1: p = 8n+5 => 8n+5 = \(x^2 + (2k-1)^2\) for different values of n and k rearranging we get \(x^2 = 4(k-k^2+1+2n)\) => \(x^2\) is a mulitple of 4 and x must be a multiple of 2 Suff. x is not divisible by 4 ( if it is \(x^2\) should be a multiple of 16)

Stat2: (x-y) = 3 or x-(2k-1)= 3 x=2(k+1) => x will be divisible by 4 depending on the value of k

Given p,x,y>0; y is odd - of the form (2k-1); p = \(x^2+y^2\) Is x divisible by 4? Stat1: p = 8n+5 => 8n+5 = \(x^2 + (2k-1)^2\) for different values of n and k rearranging we get \(x^2 = 4(k-k^2+1+2n)\) => \(x^2\) is a mulitple of 4 and x must be a multiple of 2 Suff. x is not divisible by 4 ( if it is \(x^2\) should be a multiple of 16)

Stat2: (x-y) = 3 or x-(2k-1)= 3 x=2(k+1) => x will be divisible by 4 depending on the value of k

Insuff.

IMO A

great explanation _________________

Salaries are low in recession. So, working for kudos now.