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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

IMO C.

n^2+n = n(n+1) is not divisible by 3 => n-1 is divisible by 3 IF N is no equal to 0,1,-1 else this wont hold true, thus not sufficient.

3n +5 >= k+8 => 3n >= k+3 , take k = 3m as k is positive multiple of 3

=> 3n>=3m+3 => n >= m+1 => n>1 Not sufficient.

But if we combine the both then n-1 is divisible by 3 when n>1 Thus C

n^2+n = n(n+1) is not divisible by 3 => n-1 is divisible by 3 IF N is no equal to 0,1,-1 else this wont hold true, thus not sufficient.

I guess you overlooked some facts,

Let me try to explain them with examples,

Say, n=0 then n(n+1) = 0 -> which is divisble by 3 and hence the st 1 is not valid for this example Now let n=1 then n(n+1) - > 2 which is not diviable by 3 but then n-1 = 0 which is divisble by 3 Now let n=-1 then n(n+1) = 0 which is again divisble y 3 and hence St 1 does not hold true for this example as well.

I completely understand how statement 1 is sufficient, but am going to have to review statement 2 further to understand why it is not sufficient. I understand the simple math just not the explanation that follows.

Good question though. Had me thinking. Thank you very much.

Since we are told in Statement (1) that the product n^2+n 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.

Since we are told in Statement (1) that the product n^2+n 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.

I dont buy this ...

Neither.

Here is my logic:

We know this from the question stem: n = Set of all integers = {..., -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, ... }

1) n^2 + n is not divisible by 3

We now know: n = {... -11, -8, -5, -2, 1, 4, 7, 10, ...} <--- very clear pattern here

We are interested in n-1 (but only from the above set, which meet our condition imposed on 1) n - 1 = {..., -12, -9, -6, -3, 0, 3, 6, 9, ... }

Let's check these against what ware testing for, are these divisible by three? Very clearly, yes.

Since we are told in Statement (1) that the product n^2+n 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. I dont buy this ...

Thanks Sudhir. Please notify MGMAT. Product of three consecutive integers must be divisible by 3 irrespective of 0, -ves or +ves.

(n-1)n(n+1) must be divisible by 3. n(n+1): Not Divisible (n-1): must be divisible
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Can experts say the final word regarding option (1)? I wonder whether A is sufficient

Given that n is an integer, is n — 1 divisible by 3?

(1) n^2 + n is not divisible by 3 --> n(n+1) is not divisible by 3 --> neither n nor n+1 is divisible by 3. Now, n-1, n and n+1 are three consecutive integers, thus one of them must be divisible by 3, so if n and n+1 are NOT, then n-1 must be. Sufficient.

(2) 3n +5 >= k+8 , where k is a positive multiple of 3. Not sufficient.

Re: Given that n is an integer, is n 1 divisible by 3? (1) n^2 + [#permalink]

Show Tags

02 Jun 2013, 12:25

Bunuel wrote:

LalaB wrote:

Can experts say the final word regarding option (1)? I wonder whether A is sufficient

Given that n is an integer, is n — 1 divisible by 3?

(1) n^2 + n is not divisible by 3 --> n(n+1) is not divisible by 3 --> neither n nor n+1 is divisible by 3. Now, n-1, n and n+1 are three consecutive integers, thus one of them must be divisible by 3, so if n and n+1 are NOT, then n-1 must be. Sufficient.

(2) 3n +5 >= k+8 , where k is a positive multiple of 3. Not sufficient.

Can experts say the final word regarding option (1)? I wonder whether A is sufficient

Given that n is an integer, is n — 1 divisible by 3?

(1) n^2 + n is not divisible by 3 --> n(n+1) is not divisible by 3 --> neither n nor n+1 is divisible by 3. Now, n-1, n and n+1 are three consecutive integers, thus one of them must be divisible by 3, so if n and n+1 are NOT, then n-1 must be. Sufficient.

(2) 3n +5 >= k+8 , where k is a positive multiple of 3. Not sufficient.

Answer: A.

Hope it's clear.

Can you kindly explain why B is not sufficient...

Sure.

(2) says that 3n +5 >= k+8 , where k is a positive multiple of 3 --> k=3x, for some positive integer x --> \(3n +5\geq{3x+8}\) --> \(3n-3\geq{3x}\) --> \(n-1\geq{x}\). So, basically we just have that n-1 is greater or equal to some positive integer x, thus it may or may not be a multiple of 3.

Re: Given that n is an integer, is n 1 divisible by 3? (1) n^2 + [#permalink]

Show Tags

03 Sep 2013, 05:12

Quote:

(2) says that 3n +5 >= k+8 , where k is a positive multiple of 3 --> k=3x, for some positive integer x --> \(3n +5\geq{3x+8}\) --> \(3n-3\geq{3x}\) --> \(n-1\geq{x}\). So, basically we just have that n-1 is greater or equal to some positive integer x, thus it may or may not be a multiple of 3.

Just to go a bit further on this, in what case would \(n-1\) be divisible by \(3\)? Say if you ended up with \(n-1\geq{3x}\) it would still be insufficient, is not it so? The fact that we have \(\geq\) seems to necessarily mean that \(n-1\) may not necessarily be divisible because we have so many options, or I am missing the point here?
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There are times when I do not mind kudos...I do enjoy giving some for help

n can not be 0 as o is multiple of every number and not and factor of ant number. OA seems wrong it should be A

The OA is correct.

For (1) n cannot be 0, because in this case n^2 + n = 0, which IS divisible by 3, so it would contradict this statement. But how this changes the answer? Please re-read the solutions above.
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