Is n-1 divisible by 3?

[mission800]

Given that n is an integer, is n — 1 divisible by 3?
(1) n^2 + n is not divisible by 3
(2) 3n+5 >= k+8, where k is a positive multiple of 3



I have both OA and Explanation but they're not clear to me.
Pls try to explain statement (1) in detail, i'll post the OA and explanation soon.

[rajatmeh]

Using (1)
=========
n^2 + n = n (n+1)
n^2 + n is not divisible by 3
==> neither n is divisible by 3 nor n+1 is divisible by 3

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12

If n and n+1 are not divisible by 3 means n-1 HAS TO BE divisible by 3
Easy way to explain is, by looking at example from above numbers
if n=4 n+1=5 n-1=3, divisible by 3
if n=1 n+1=2 n-1=0, divisible by 3

Sufficient

Using (2)
=========
3n+5 >= k+8, where k is a positive multiple of 3
Assume k = 3t
3n+5 >= 3t+8
n > = t+1
n-1>=t
==> t can further be multiple of 3 or it cannot be
Not sufficient.

Answer (A)

[12rk34]

(1) n*(n+1) is not divisible by 3.
This is possible if n=3k+1 where k is any integer (k=-3, -2, -1, 0, 1, 2...).
n*(n+1) = (3k+1)(3k+2)
= 9K^2 + 9k + 2 is not divisible by 3 since last term i.e. 2 is not divisible by 3.
Obviously n-1 =3k is divisible by 3.
Hence A.

[nithesh]

I have realised that understanding the principles behind the relations between numbers saves your time as compared to direct calculation.

Just a different way of looking at this question

Starting from 1 very third number is a multiple of 3. This means when you consider any 3 consecutive numbers, one (and only one) of them is a multiple of 3. (This is true for - to + infinity)
As the first equation says n & (n+1) are not multiples of 3, (n-1) & (n+1) have ot be multiples of 3. - SUFFICIENT

3n+5 >= k+8
As k is a multiple of 3, k+3 is also a multiple of 3 and as we don't have to find out the vaue of n or k, we can simply replace k+3 by k.
Hence:
3n+5 >= k+5
i.e. 3n >= k
This does not say anything at all!!! - NOT SUFFICIENT

[mission800]

ok. i did the same thing, so hopefully the OA/explanation is wrong.
I'll post it, pls take a look:
(this is from Manhattan-Tough Quant document i have. maybe its official, or maybe its compiled by someone)

OA-C
Statement (1) gives us information about n^2 + n, which can be rewritten as the product of two consecutive integers as follows: n(n+1)
Since the question asks us about n — 1, we can see that we are dealing with three consecutive integers: n — 1, n, and n + 1 .
By definition, the product of consecutive nonzero integers is divisible by the number of terms. Thus the product of three consecutive nonzero integers must be divisible by 3.
Since we are told in Statement (1) that the product n(n+1) is not divisible by 3, we know that neither n nor n + 1 is divisible by 3. Therefore it seems that n — 1 must be divisible by 3.
However, this only holds if the integers in the consecutive set are nonzero integers. Since Statement (1) does not tell us this, it is not sufficient.
Statement (2) can be rewritten as follows:
3n+5>=k+8
n >= k/3 + 1
Given that k is a positive multiple of 3, we know that n must be greater than or equal to 2. This tells us that the members of the consecutive set n — 1, n, n + 1 are nonzero integers.
By itself, this information does not give us any information about whether n — 1 is divisible by 3. Thus Statement (2) alone is not sufficient.
When both statements are taken together, we know that the members of the consecutive set n — 1, n, n + 1 are nonzero integers and that neither n nor n + 1 is divisible by 3. Therefore, n — 1 must be divisible by 3.
The correct answer is C: both statements together are sufficient but neither statement alone is sufficient to answer the question.

i think (1) should be sufficient, comments ?

[nithesh]

Is zero an integer?
Is -3 a multiple of 3?

[mission800]

-yes (anything thats not a fraction)
-yes 3 X (-1)

[rajatmeh]

I kept 0 and negative integers in mind while solving the question.
I feel that even 0 is divisible by 3. I am not convinced with the solution explanation given.

[mission800]

yes, i was in the same dilemma, particularly because i dont expect the solution of this standard to be wrong, but i guess in this case it just is.

and yes, 0 is definitely divisible by every number (except 0)

[zapped]

if we chose n=-1 then (1) = -1(-1+1) = -1 which is not divisible by 3.
If i take n-1, then n-1=-1-1= -2 which is not divisible by 3.
However, if I take n=1, then 1(1+1)= 2 (not divisible by3), but n-1= 0 which is divisible. So, I kindda agree with the OG solution. (n-1), n and n+1 should be positive integers for the conditions to satisfy.
Let me know if I misinterpreted something here.