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If p and n are positive integers and p>n, what is the remainder when p^2 - n^2 is divided by 15?

First of all \(p^2 - n^2=(p+n)(p-n)\).

(1) The remainder when p + n is divided by 5 is 1. No info about p-n. Not sufficient.

(2) The remainder when p - n is divided by 3 is 1. No info about p+n. Not sufficient.

(1)+(2) "The remainder when p + n is divided by 5 is 1" can be expressed as \(p+n=5t+1\) and "The remainder when p - n is divided by 3 is 1" can be expressed as \(p-n=3k+1\).

Multiply these two --> \((p+n)(p-n)=(5t+1)(3k+1)=15kt+5t+3k+1\), now first term (15kt) is clearly divisible by 15 (r=0), but we don't know about 5t+3k+1. For example t=1 and k=1, answer r=9 BUT t=7 and k=3, answer r=0. Not sufficient.

OR by number plugging: if \(p+n=11\) (11 divided by 5 yields remainder of 1) and \(p-n=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(p-n)=11\) and remainder upon division 11 by 15 is 11 BUT if \(p+n=21\) (21 divided by 5 yields remainder of 1) and \(p-n=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(p-n)=21\) and remainder upon division 21 by 15 is 6. Not sufficient.

Hi Bunuel, According to my understanding ans should be c.. given (p+n)/5 = rem(1) (p-n)/3= rem(1) so (p^2 - n^2)/15 = (p+n)/5 * (p-n)/3... so remainder will be equal to 1*1 = 1

Hi Bunuel, According to my understanding ans should be c.. given (p+n)/5 = rem(1) (p-n)/3= rem(1) so (p^2 - n^2)/15 = (p+n)/5 * (p-n)/3... so remainder will be equal to 1*1 = 1

please correct me where I am wrong.

Red part is not correct.

There are both algebraic and number plugging approaches in my previous post showing that answer is E. Yuo can check it yourself:

If \(p=6\) and \(n=5\) then \(p+n=11\) (11 divided by 5 yields remainder of 1) and \(p-n=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(p-n)=11\) and remainder upon division 11 by 15 is 11

If \(p=11\) and \(n=10\) then \(p+n=21\) (21 divided by 5 yields remainder of 1) and \(p-n=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(p-n)=21\) and remainder upon division 21 by 15 is 6.

Two different answers. Not sufficient.
_________________

Re: If p and n are positive integers and p > n, what is the rema [#permalink]

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09 Aug 2013, 08:07

Hi Bunuel,

I understood your approach for this problem. However , would like to have your opinion why the below solution as given in the older post is wrong?

so (p^2 - n^2)/15 = (p+n)/5 * (p-n)/3... so remainder will be equal to 1*1 = 1

Please advise as to what was wrong in this solution in detail.

Rgds, TGC!
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

Re: If p and n are positive integers and p > n, what is the rema [#permalink]

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20 Sep 2015, 05:23

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Re: If p and n are positive integers and p > n, what is the rema [#permalink]

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01 Oct 2016, 01:59

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).

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Re: If p and n are positive integers and p > n, what is the rema [#permalink]

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01 Oct 2016, 21:27

sudhanshushankerjha wrote:

Hi Bunuel, According to my understanding ans should be c.. given (p+n)/5 = rem(1) (p-n)/3= rem(1) so (p^2 - n^2)/15 = (p+n)/5 * (p-n)/3... so remainder will be equal to 1*1 = 1

please correct me where I am wrong.

please see, If p=14, n=7 then p+n=21 mean (p+n)/5, remainder 1 p-n=7,means( p-n)/3, remainder 1

then p^2-n^2= 196-49=147 divided by 15 remainder 12. will not be able to find using 1 & 2. Hence answer is E

gmatclubot

Re: If p and n are positive integers and p > n, what is the rema
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01 Oct 2016, 21:27

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