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If p and n are positive integers and p > n, what is the rema

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

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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?

(1) The remainder when p + n is divided by 5 is 1.
(2) The remainder when p - n is divided by 3 is 1.
[Reveal] Spoiler: OA

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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.

Answer: E.
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Re: DS8 [#permalink]

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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.
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Re: DS8 [#permalink]

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New post 04 Oct 2010, 14:05
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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.


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

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

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New post 04 Aug 2013, 01:26
If p and n are positive integers and p>n, what is the remainder when \(p^2 - n^2\) is devided by 15?

1) The remainder when p+n is devided by 5 is 1.
2) The remainder when p-n is devided by 3 is 1.
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Last edited by Zarrolou on 04 Aug 2013, 01:28, edited 1 time in total.
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Re: If p and n are positive integers and p > n, what is the rema [#permalink]

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New post 09 Aug 2013, 09:07
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.

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

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Re: If p and n are positive integers and p > n, what is the rema   [#permalink] 20 Sep 2015, 06:23
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