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Divisibility by 3

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by knight247 » Tue Feb 14, 2012 8:54 am
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.
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by rijul007 » Tue Feb 14, 2012 9:23 am
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.
Actually Zero is divisible by 3.

When N divided by X gives a remainder 0, we say N is divisible by X.

Similarly 0 divided by 3 would give us a remainder 0. Hence, 0 is divisible by 3.
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by Brent@GMATPrepNow » Tue Feb 14, 2012 9:26 am
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
Brent Hanneson - Creator of GMATPrepNow.com
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by LalaB » Tue Feb 14, 2012 9:37 am
I didnt get u .why 0 cant be divisible by 3? 0/3=0



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


stmt 1 - n²+n =n(n+1) <--consecutive numbers . so we have 3 consecutive numbers (n-1 ) n (n+1)

cases-
0 1 2
3 4 5
6 7 8

smtnt2 -
3n=k+3
3n-3=k
n-1=k/3

k is divisible by 3. let k =3 then n-1=1 not divisible by 3
let k =9 then n-1=3 divisible by 3. so insuff

oops, I am late :) . the q. is answered already
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