Divisibility by 3

knight247 wrote:If n is a positive integer, is n-1 divisible by 3?
(1)n²+n is not divisible by 6
(2)3n=k+3, where k is a positive multiple of 3

The OA is A

Here is where I have a problem with the OA.
If n=1 then n²+n=2 which isn't divisible by 6. So n=1 is a valid case. But n-1 would equal zero which obviously isn't divisible by 3. All the other values like n=4, n=7 ETC seem to meet the requirement. But n=1 clearly proves the insufficiency of (1). Hoping to get a clear explanation.

Yes, zero is definitely divisible by 3.

Here's my solution.


Target question: Is n-1 divisible by 3?

Statement 1: n²+n is not divisible by 6
This statement is testing two things.
First, there's are great rule that says: If we have n consecutive integers, then one of those integers must be divisible by n.
So, for example, if we have 5 consecutive integers, one of those numbers will be divisible by 5.
Second, it's testing our ability to see that n^2 + n = n(n+1), and that n and n+1 are two consecutive integers.

Now, for n(n+1)to be divisible by 6, it would have to be divisible by 2 and by 3
Since n and n+1 are two consecutive integers, then one must be divisible by 2, so n(n+1)is definitely divisible by 2.
However, we are told that n(n+1)is not divisible by 6, which means that neither n nor n+1 is divisible by 3.

Now recognize that n-1, n and n+1 are three consecutive integers. Since we now know that neither n nor n+1 is divisible by 3, it must be true that n-1 is divisible by 3.
So statement 1 is SUFFICIENT

Statement 2: 3n=k+3, where k is a positive multiple of 3
Let's plug in some numbers here.
case a: k=3, which means n=2, which means n-1 is not divisible by 3
case b: k=9, which means n=4, which means n-1 is divisible by 3
So statement 2 is INSUFFICIENT

The answer is A

Cheers,
Brent