Is positive integer n 1 a multiple of 3?

But n-1 isn't mentioned in the 2) so how can we say something about it? how come that if n or n+1 is div by 3 then automatically n-1 is not?..

Out of any 3 consecutive integers only 1 is a multiple of 3. We know it's either n or n + 1, therefore it's not n-1.

ahhhhh okok! now got it.. omg this is a tricky one....

Thank you Bunuel!

Hello,

What is the approximate level of this question on 800 scale?

Regards.

Check the original post for tags and other stats. It's a 700-level question.

Target question: Is positive integer n – 1 a multiple of 3?

Statement 1: n³ – n is a multiple of 3
Let's do some FACTORING
n³ – n = n(n² - 1) = n(n + 1)(n - 1)
So, statement 1 tells us that n(n + 1)(n - 1) is a multiple of 3
Notice that (n-1), n and (n+1) are 3 CONSECUTIVE integers, and we know that 1 out of every 3 consecutive integers is a multiple of 3 (e.g., 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, etc.
So, ONE of the following n, (n + 1), or (n - 1) a multiple of 3, but WHICH ONE??
It COULD be the case that (n - 1) IS a multiple of 3, OR it could be the case that (n - 1) is NOT a multiple of 3
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: n³ + 2n² + n is a multiple of 3
More FACTORING
n³ + 2n² + n = n(n² + 2n + 1)
= n(n + 1)(n + 1)
So, statement 2 tells us that n(n + 1)(n + 1) is a multiple of 3
This means that EITHER n is a multiple of 3 OR (n + 1) is a multiple of 3
Let's examine both cases:

Case a: n is a multiple of 3
If n is a multiple of 3, then n - 1 CANNOT be a multiple of 3.
Why not?
Well, we already know that 1 out of every 3 consecutive integers is a multiple of 3
So, if n is a multiple of 3, the n+3 is also a multiple of 3 AND n+6 is a multiple of 3, AND n+9 is a multiple of 3, etc
Likewise, n-3 is a multiple of 3 AND n-6 is a multiple of 3, AND n-9 is a multiple of 3, etc.
Notice that n-1 is NOT among the possible multiples of 3
So, in this case, the answer to the target question is NO, n - 1 is NOT a multiple of 3

Case b: (n + 1) is a multiple of 3
If n+1 is a multiple of 3, then n - 1 CANNOT be a multiple of 3.
We'll use the same logic we used in case a above
1 out of every 3 consecutive integers is a multiple of 3
So, if n+1 is a multiple of 3, the n+1+3 (aka n+4) is also a multiple of 3 AND n+1+6 (aka n+7) is a multiple of 3, AND n+1+9 (aka n+10)is a multiple of 3, etc
Likewise, n+1-3 (aka n-2) is a multiple of 3 AND n+1-6 (aka n-5) is a multiple of 3, AND n+1-6 (aka n-5) is a multiple of 3, etc.
Notice that n-1 is NOT among the possible multiples of 3
So, in this case, the answer to the target question is NO, n - 1 is NOT a multiple of 3

IMPORTANT: for statement 2, there are only two possible cases, and in each case, the answer to the target question is the SAME: NO, n - 1 is NOT a multiple of 3
So, it MUST be the case that n - 1 is NOT a multiple of 3
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent

Bunuel chetan2u, can you please clarify st 2 for n-1=3 (n=4)?

From (2) we get that either n or n + 1, is a multiple of 3, thus it cannot be n - 1. Hence, your choice of n - 1 does not satisfy (2) and therefore is invalid.