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If p and n are positive integers and p > n, what is the rema [#permalink]
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10 May 2010, 09:34
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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.
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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 \(p^2  n^2=(p+n)(pn)\). (1) The remainder when p + n is divided by 5 is 1. No info about pn. 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 \(pn=3k+1\). Multiply these two > \((p+n)(pn)=(5t+1)(3k+1)=15kt+5t+3k+1\), 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 \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=11\) and remainder upon division 11 by 15 is 11 BUT if \(p+n=21\) (21 divided by 5 yields remainder of 1) and \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=21\) and remainder upon division 21 by 15 is 6. Not sufficient. Answer: E.
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Hi Bunuel, According to my understanding ans should be c.. given (p+n)/5 = rem(1) (pn)/3= rem(1) so (p^2  n^2)/15 = (p+n)/5 * (pn)/3... so remainder will be equal to 1*1 = 1
please correct me where I am wrong.



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sudhanshushankerjha wrote: Hi Bunuel, According to my understanding ans should be c.. given (p+n)/5 = rem(1) (pn)/3= rem(1) so (p^2  n^2)/15 = (p+n)/5 * (pn)/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 \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=11\) 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 \(pn=1\) (1 divided by 3 yields remainder of 1) then \((p+n)(pn)=21\) 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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19 Jul 2013, 01:24



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If p and n are positive integers and p>n, what is the remain [#permalink]
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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 pn 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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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 * (pn)/3... so remainder will be equal to 1*1 = 1 Please advise as to what was wrong in this solution in detail. Rgds, TGC!
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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, 02:59
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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, 22:27
sudhanshushankerjha wrote: Hi Bunuel, According to my understanding ans should be c.. given (p+n)/5 = rem(1) (pn)/3= rem(1) so (p^2  n^2)/15 = (p+n)/5 * (pn)/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 pn=7,means( pn)/3, remainder 1 then p^2n^2= 19649=147 divided by 15 remainder 12. will not be able to find using 1 & 2. Hence answer is E




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