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For 1, for following values of n, \(n^3 - n\) becomes multiple of 3 : 2,3,4,5,6,7,8....
However, by plugging these values in n-1, we have different results. thus, not sufficient
For 2, following values of n gets the equation as a multiple of 3 : 2,3,5,6,7,8.... Its again not sufficient.
Combining 1 & 2, we again can't be conclusive. Thus,answer should be E.
Please write where am I wrong?
Is positive integer n – 1 a multiple of 3?
(1) n^3 – n is a multiple of 3 --> \(n^3-n=n(n^2-1)=(n-1)n(n+1)=3q\). Now, \(n-1\), \(n\), and \(n+1\) are 3 consecutive integers and one of them must be multiple of 3, so no wonder that their product is a multiple of 3. However we don't know which one is a multiple of 3. Not sufficient.
(2) \(n^3 + 2n^2+ n\) is a multiple of 3 --> \(n^3 + 2n^2+ n=n(n^2+2n+1)=n(n+1)^2=3p\) --> so either \(n\) or \(n+1\) is a multiple of 3, as out of 3 consecutive integers \(n-1\), \(n\), and \(n+1\) only one is a multiple of 3 then knowing that it's either \(n\) or \(n+1\) tells us that \(n-1\) IS NOT multiple of 3. Sufficient.
Re: Is positive integer n 1 a multiple of 3? [#permalink]
27 Jun 2015, 04:19
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