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if n is a positive integer and r is the remainder when

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if n is a positive integer and r is the remainder when [#permalink]

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New post 02 Aug 2008, 06:53
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if n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24. what is the value of r?

(1) n is not divisible by 2
(2) n is not divisible by 3

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Re: DS: remainder (GMATPREP 1) [#permalink]

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New post 02 Aug 2008, 07:17
selvae wrote:
Answer is C


You are right, selvae.

Could you provide explanation?

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Re: DS: remainder (GMATPREP 1) [#permalink]

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New post 02 Aug 2008, 07:37
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n-1 and n+1 are consecutive even numbers (since n is not a factor of 2 and 3 so n is odd)

Difference between n-1 and n+1 is 2, so either of the one will be divisible by 2 and other will be divisible by 4,
since they are consecutive even numbers.

Also since 3 is not a factor of n, either n-1 or n+1 should be divisible by 3,

so we got 2*4*3 = 24.

so we will get reminder as 0 always.

try with number 5 which is the minimum possible number....

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Re: DS: remainder (GMATPREP 1) [#permalink]

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New post 02 Aug 2008, 07:39
selvae wrote:
n-1 and n+1 are consecutive even numbers (since n is not a factor of 2 and 3 so n is odd)

Difference between n-1 and n+1 is 2, so either of the one will be divisible by 2 and other will be divisible by 4,
since they are consecutive even numbers.

Also since 3 is not a factor of n, either n-1 or n+1 should be divisible by 3,

so we got 2*4*3 = 24.

so we will get reminder as 0 always.

try with number 5 which is the minimum possible number....


Wow, nice one. Thanks!!

+1 for you

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Re: DS: remainder (GMATPREP 1) [#permalink]

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New post 02 Aug 2008, 11:12
judokan wrote:
if n is a positive integer and r is the remainder when (n-1)(n+1) is divided by 24. what is the value of r?

(1) n is not divisible by 2
(2) n is not divisible by 3


consider (1) => say n=3 => n^2-1=8 => remainder 8
also say n=5 => remainder =0 => insufficient

consider (2) => say n=2 => remainder =3
say n=4 => remainder =15 => insufficient

consider (1) and (2) => n is not div by 2 and 3 ,also not div by 6,4 etc
hence say n is div 5 => n=5 => remainder=0
n=10 => remainder =6 => insufficient

IMO E
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Re: DS: remainder (GMATPREP 1) [#permalink]

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New post 02 Aug 2008, 17:55
spriya wrote:
judokan wrote:
selvae wrote:
n-1 and n+1 are consecutive even numbers (since n is not a factor of 2 and 3 so n is odd)

Difference between n-1 and n+1 is 2, so either of the one will be divisible by 2 and other will be divisible by 4,
since they are consecutive even numbers.

Also since 3 is not a factor of n, either n-1 or n+1 should be divisible by 3,

so we got 2*4*3 = 24.

so we will get reminder as 0 always.

try with number 5 which is the minimum possible number....


OOPs good explanation again i messed up taking n=10 as example

selvae,where do u practice Quant from :|
Wow, nice one. Thanks!!

+1 for you


I think the key of this one is first to list out the numbers for n,

1,2,3,4,5,6,7,8,9,10 ...
and then eliminate, as n is not a multiple of 2,
1,3,5,7,9,
then you will see the remainder changing among different value of n.
so not siufficient

if we add (2)
then the value of n could be
1,5,7,11,13,17
you will see all these value n will give (n+1)(n-1) divisbile by 24

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Re: DS: remainder (GMATPREP 1)   [#permalink] 02 Aug 2008, 17:55
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if n is a positive integer and r is the remainder when

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