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.
Divisibility by 3
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- rijul007
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Actually Zero is divisible 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.
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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- Brent@GMATPrepNow
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Yes, zero is definitely divisible 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.
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
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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
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
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