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 is a positive odd integer, what is the remainder when p is divided by 4 ?

(1) When p is divided by 8, the remainder is 5 --> \(p=8q+5=(8q+4)+1=4(2q+1)+1\) --> so the remainder upon division of p by 4 is 1 (since first term is divisible by 4 and second term yields remainder of 1 upon division by 4). Sufficient.

(2) p is the sum of the squares of two positive integers --> since p is an odd integer then one of the integers must be even and another odd: \(p=(2n)^2+(2m+1)^2=4n^2+4m^2+4m+1=4(n^2+m^2+m)+1\) --> the same way as above: the remainder upon division of p by 4 is 1 (since first term is divisible by 4 and second term yields remainder of 1 upon division by 4). Sufficient.

If p is a positive odd integer, what is the remainder when p [#permalink]

Show Tags

26 Jul 2012, 07:05

I am not sure if this is discussed, tried to find it but couldn't find easily.

If p is a positive odd integer, what is the remainder when p is divided by 4 ? (1) When p is divided by 8, the remainder is 5. (2) p is the sum of the squares of two positive integers.

I need help, here i thought the answer was A but the official answer given is D. Here's how i did it

(1) P = 8Q + 5, so for all values of Q starting from 1, 4 divided by P results in remainder 1. So sufficient (2) P = X^2+ Y^2, ex X=Y=1, P= 2, remainder 2. ex X=1,Y=2, P=5, remainder 1. Hence Not Sufficient.

I am not sure if this is discussed, tried to find it but couldn't find easily.

If p is a positive odd integer, what is the remainder when p is divided by 4 ? (1) When p is divided by 8, the remainder is 5. (2) p is the sum of the squares of two positive integers.

I need help, here i thought the answer was A but the official answer given is D. Here's how i did it

(1) P = 8Q + 5, so for all values of Q starting from 1, 4 divided by P results in remainder 1. So sufficient (2) P = X^2+ Y^2, ex X=Y=1, P= 2, remainder 2. ex X=1,Y=2, P=5, remainder 1. Hence Not Sufficient.

Confused?

Merging similar topics.

Please hide OA under the spoiler.
_________________

Re: If p is a positive odd integer, what is the remainder when p [#permalink]

Show Tags

26 Jul 2012, 07:11

Hi Summer101

For St 2: U have to make P an odd Integer. If u select X = Y = 1, Then P become 2 i.e. even. Thats why D is the correct answer.
_________________

Regards SD ----------------------------- Press Kudos if you like my post. Debrief 610-540-580-710(Long Journey): http://gmatclub.com/forum/from-600-540-580-710-finally-achieved-in-4th-attempt-142456.html

Re: If p is a positive odd integer, what is the remainder when p [#permalink]

Show Tags

25 Jan 2013, 20:43

number plugging is the fastest approach for remainder problems.. because there emerges a definite pattern.. that can make the stem sufficient.
_________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

Re: If p is a positive odd integer, what is the remainder when p [#permalink]

Show Tags

31 Jan 2013, 04:03

OK.. Let me quote here something..

Let's consider x=1 and y=2... square of two number is 5 (a positive odd integer) and leaves remainder 5 when divided by 8 Again..consider x=1 and y = 6, square of two numbers is 37 (a positive odd integer) and leaves remainder 5 when divided by 8 hence, both the statements are either sufficient to ans this problem..

Re: If p is a positive odd integer, what is the remainder when p [#permalink]

Show Tags

04 Sep 2013, 10:12

D

1) Is sufficient 5,13 all give 1 as remainder. 2) sum of squares of any two positive integers, but one of them has to be odd and other an even number because p is an odd integer. So consider any pair 3,2 (3^2+2^2) (9+4)/4 1 as remainder. Or 5,2 =>29/4 =>1 as remainder.
_________________

--It's one thing to get defeated, but another to accept it.

The question asks what will be the remainder when P is divided by 4.

Statement 1 says: p=8n+5…..which is when P is divided by 8 we get remainder 5. Now from here on you can do two things. 1. think logically 8n is divided by 4 so we won't have any remainder if we divide 8n by 4 but if we divide 5 by 4 we always get remainder 1. The value of N doesn't really matter because 8N will always be divisible by 4 and 5 will always give remainder 1. 2. plug in number and see what happens. If we put n=1,2,3,4 or so on….we get 13,21,29 respectively. Now in each case we get remainder 1.

So statement 1 is sufficient. Answer should be either A or D. So cross out B C and E.

Statement 2 says: p is sum of square of two integers just plug in numbers and see you will always get remainder 1. So statement 2 is sufficient and Answer is D.

Re: If p is a positive odd integer, what is the remainder when p [#permalink]

Show Tags

28 Sep 2014, 06:29

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.
_________________

If p is a positive odd integer, what is the remainder when p [#permalink]

Show Tags

20 Jul 2015, 09:23

Can you please explain why the other number has to be even as p is odd? When we take two odd numbers ie {(1,3),(3,5),(5,7)} etc.. the sum of squares are (10,34,74) and all those give a remainder of 2. Please do explain.

Re: If p is a positive odd integer, what is the remainder when p [#permalink]

Show Tags

21 Jul 2015, 00:46

1

This post received KUDOS

chandrae wrote:

Can you please explain why the other number has to be even as p is odd? When we take two odd numbers ie {(1,3),(3,5),(5,7)} etc.. the sum of squares are (10,34,74) and all those give a remainder of 2. Please do explain.

Hi Chandrae,

According to the question, p needs to be a positive odd integer.

If p is a positive odd integer, what is the remainder when p is divided by 4 ?

(1) When p is divided by 8, the remainder is 5. (2) p is the sum of the squares of two positive integers.

In statement 2, p can only be odd if one of the squares is odd and the other is even. ( Even + Even = Even, Odd + Odd = Even and Even + Odd = Odd).

Therefore, the cases you mentioned (in which the sum of squares is even) can not be considered as eligible values of p as per the question.

Re: If p is a positive odd integer, what is the remainder when p [#permalink]

Show Tags

06 Jan 2016, 11:40

1

This post received KUDOS

BANON wrote:

If p is a positive odd integer, what is the remainder when p is divided by 4 ?

(1) When p is divided by 8, the remainder is 5. (2) p is the sum of the squares of two positive integers.

Got it wrong, was happy to see that Statement 1 is sufficient and rushed on Statement 2..)))

(1) p=8k+5: 5, 13, 21, 29 etc. we have a valid pattern here and each time a remainder of 1. Sufficient (2) p is the sum of the squares of two positive integers -> the least possible number is \(1^2+2^2=5\) , 13, 17, 25, 29 and each time we have a remainder of 1 (actually almost the samt thing as above) Sufficient

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 p is a positive odd integer, what is the remainder when p is divided by 4 ?

(1) When p is divided by 8, the remainder is 5. (2) p is the sum of the squares of two positive integers.

In the original condition, there is 1 variables(p), which should match with the number of equation. So you need 1 equation. For 1) 1euquation, for 2) 1 equation, which is likely to make D the answer. In 1), the remainder from p=8m+5=4(2m+1)+1 is 1, which is unique and sufficient. In 2), from p=odd=a^2+b^2, either a or b should be an even integer and the other should be an odd integer. In case of even^2+odd^2, even^2 is always divided by 4. For odd^2, from 1^2=1, 3^2=9, 5^2=25, 7^2=49, 9^2=81, all of them are divided by 4 and the remainder is 1, which is unique and sufficient. Therefore, the answer is D.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...