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

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Director
Joined: 30 Nov 2006
Posts: 591

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

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01 Jun 2007, 14:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

(a) The remainder when p+n is divided by 5 is 1
(b) The remainder when p-n is divided by 3 is 1

I will provide the OA soon.

Here is how I started to appraoch the problem, but then got stuck:

p^2-n^2 = (p-n)(p+n)
(p-n)(p+n) = 15Q + R [ Q is the quotient and R is the asked remainder ]

(a) p+n = 5Q + 1
(b) p-n = 3Q + 1

I thought since 15 has 3 and 5 as its factors, the answer will be C. Obviously, I was wrong ! I dunno why
[/u]

Kudos [?]: 314 [0], given: 0

VP
Joined: 08 Jun 2005
Posts: 1144

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02 Jun 2007, 05:21
(P^2-n^2)/15

what is the remainder ?

(P^2-n^2)/15 = (p+n)(p-n)/5*3 = (p+n)/5*(p-n)/3

statement 1

(p+n)/5 will yield remainder of 1.

since (p-n)/3 is unknown

insufficient

statement 2

(p-n)/3 will yield remainder of 1.

since (p+n)/5 is unknown

insufficient

statements 1&2

since we can use the multiplication of two remaniders to find the joined remiander (and vice versa) only when the base is qeual.

bold = base

e.g (7/3*5/3) = (1*2)/3 = 2 or (16/7*25/7) = 8/7 = 1

but not when (12/5*7/3) = (2*1)/??? = ???

then:

(p-n)/3*(p+n)/5 = cannot be solved since the base is not equal (3,5) !!

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Intern
Joined: 04 Feb 2007
Posts: 37

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02 Jun 2007, 08:01
p^2-n^2 = (p+n)*(p-n)

1)(p+n)=(5*p+1) ...insuff.

2)(p-n)=(3*q+1) ...insuff.

Together,

p^2-n^2 = (p+n)*(p-n) = (5*p+1) * (3*q+1)
=(15*pq+5*p+3*q+1)
=insufficient as p & q are still unknown. (E)

Kudos [?]: 2 [0], given: 0

02 Jun 2007, 08:01
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