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# If n is a positive integer, is n^3-n divisible by 4? I)

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If n is a positive integer, is n^3-n divisible by 4? I) [#permalink]

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14 Jan 2004, 17:06
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If n is a positive integer, is n^3-n divisible by 4?

I) n=2k+1, where k is an integer
II) n^2+n is divisible by 6

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14 Jan 2004, 18:20
n^3-n

Do you mean (n^3)-n or n^(3-n)?

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14 Jan 2004, 20:39
Not sure but my answer would be (A)

Here is my approach :
Note : I am assuming that the expression is (n)^3-n

n^3-n = n(n^2-1) = n(n+1)(n-1)

1) n=2k+1
So n^3-n = (2k+1)(2k+1+1)(2k+1-1)
= (2k+1)(2k+2)(2k)
= 4k(2k+1)(k+1) = 4m
hence divisible by 4 for all values of m

2) n^2+n is divisible by 6
n^2+n = n(n+1) = 6p
Now n^3-n = n(n+1)(n-1) = 6p(n-1)
This is not divisible by 4 for all values of p and n
For e.g., if p=1 and n=2, this doesn't hold good

Last edited by giddi77 on 15 Jan 2004, 11:26, edited 1 time in total.

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14 Jan 2004, 21:02
I am assuming (n^3)-n and answer I get is A
The way I approached this problem was by picking a few integers and try see if it works.
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Best Regards,

Paul

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14 Jan 2004, 21:12
Paul wrote:
I am assuming (n^3)-n and answer I get is A
The way I approached this problem was by picking a few integers and try see if it works.

I agree. There is no point in trying to prove for a generic equation

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15 Jan 2004, 06:09
giddi77 wrote:
Paul wrote:
I am assuming (n^3)-n and answer I get is A
The way I approached this problem was by picking a few integers and try see if it works.

I agree. There is no point in trying to prove for a generic equation

Well.. I guess the point would be to make sure it works for all integers and not for just the ones you've picked.

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15 Jan 2004, 06:10
stoolfi wrote:
n^3-n

Do you mean (n^3)-n or n^(3-n)?

Sorry about that, I thought it was a convention in this forum that n^3-n meant (n^3)-n

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15 Jan 2004, 06:10
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