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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?

(1) n^3-n = n(n^2-1) = n(n+1)(n-1) This number will always be a multiple of three ... so not sufficient to answer our question

(2) n^3 + 2n^2+ n = n(n+1)^2 If this is a multiple of 3, then either n is a multiple of 3 or (n+1) is a multiple of 3. In either case, n-1 cannot be a multiple of 3. Hence this is sufficient

Answer should be (b)

In your solution above, notice that for all the n's in case (2) n-1 is not a multiple of 3. _________________

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.

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...