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# If n is a positive integer, is n3 n divisible by 4? 1. n =

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SVP
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If n is a positive integer, is n3 n divisible by 4? 1. n = [#permalink]

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24 Feb 2008, 17:59
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147. If n is a positive integer, is n3 – n divisible by 4?

1. n = 2k + 1
2. n2 + n is divisible by 6
[Reveal] Spoiler: OA

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Re: If n is a positive integer, is n3 n divisible by 4? 1. n = [#permalink]

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24 Feb 2008, 19:03
Taking too long on this one ... have it down to C or E

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Re: If n is a positive integer, is n3 n divisible by 4? 1. n = [#permalink]

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24 Feb 2008, 20:30
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i get A for this one... a lil rusty..

here is how

stem asks (n-1)*n*(n+1) is divisible by 4?

well basically we know the above is divisible by 3..

1) n=2K+1 basically says n=odd
if n is odd, yes the above is divisible by 4..sufficient

2) n(n+1)=6
basically n can be 2 or 3..or n can be 3 or 4.. if n=2 then the stem is not divisible by 4..if n=3 it is divisible by 4..insuff..

A it is..

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Re: If n is a positive integer, is n3 n divisible by 4? 1. n = [#permalink]

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24 Feb 2008, 20:33
Is the question complete? please use "^" for exponents, if any.

Assuming n3-n = n^3 - n

1. suff. whether n = 2^k + 1 or n = 2*k + 1, n is always odd.
2. insuff. assuming the equation is n^2 + n, if n is divisible by 6 it doesn't meant n is divisible by 4.

So A.

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Re: If n is a positive integer, is n3 n divisible by 4? 1. n = [#permalink]

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24 Feb 2008, 22:42
sreehari wrote:
Is the question complete? please use "^" for exponents, if any.

Assuming n3-n = n^3 - n

1. suff. whether n = 2^k + 1 or n = 2*k + 1, n is always odd.
2. insuff. assuming the equation is n^2 + n, if n is divisible by 6 it doesn't meant n is divisible by 4.

So A.

You are right!

And OA is A, thank you!
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Re: If n is a positive integer, is n3 n divisible by 4? 1. n = [#permalink]

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25 Feb 2011, 23:25
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FN wrote:
i get A for this one... a lil rusty..

here is how

stem asks (n-1)*n*(n+1) is divisible by 4?

well basically we know the above is divisible by 3..

1) n=2K+1 basically says n=odd
if n is odd, yes the above is divisible by 4..sufficient

2) n(n+1)=6
basically n can be 2 or 3..or n can be 3 or 4.. if n=2 then the stem is not divisible by 4..if n=3 it is divisible by 4..insuff..

A it is..

I like the approach you have taken here to translate n^3 - n to (n-1)*n*(n+1). It makes A a clear choice. I went a longer route of seeing whether n(n^2 - 1) is divisible by 4 i.e. whether either n on its own or n^2 - 1 was divisible by 4. It took WAY longer than 2 mins to reach the conclusion. Simplifying n*(n^2 -1) one step further makes total sense.
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Re: If n is a positive integer, is n3 n divisible by 4? 1. n = [#permalink]

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26 Feb 2011, 02:51
This question is discussed here: og-ds-170arithmetic-109941.html
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Re: If n is a positive integer, is n3 n divisible by 4? 1. n = [#permalink]

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04 Nov 2017, 01:28
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Re: If n is a positive integer, is n3 n divisible by 4? 1. n =   [#permalink] 04 Nov 2017, 01:28
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# If n is a positive integer, is n3 n divisible by 4? 1. n =

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