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

Question

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

Bunuel · Accepted Answer

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.

Span · Answer

1 - neither n or n+1 are divisible by 3. Thus, n-1 must be divisible by 3, since every 3rd integer is divisible by 3. SUFF

2 - 3n + 5 >= k + 8

3n - 3 >= k

3(n-1) >= k

Because we know k is divisible by 3, but not 9 (3x3), n-1 could or could not be divisible by 3. INS

Answer A

cipher · Answer

A,

St1  n^2 + n = n* (n +1)  is not divisible by 3

which means neither n nor n+1 is divisble by 3 hence n-1 will be divisble by 3

St2 - 3n +5 >= k +8 = > 3n >= K

Now n can be a divisble by 3 or not

for .e g take n = 3 and k = 2 then n is divisble by 3 but if n =4 and k = 2 then n is not.

OA please?

gurpreetsingh · Answer

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

cipher · Answer

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.

Cheers,

skipjames · Answer

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.

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

Testluv · Answer

(1) states that n^2 + n is not divisible by 3.

Or, n*(n+1) is not divisible by 3.

Thus, neither "n" nor "n+1" is a multiple of 3.

Since every 3rd integer on the number line is a multiple of 3, this means that "n+2" is a multiple of 3. And, again, since every 3rd number is a mutiple of 3, (n+2)-3 or "n-1" is a multiple of 3.
Sufficient.

(2) is an inequality. Even if it said:

n-1>= a multiple of 3 (or n-1>= non-multiple of 3)

it would still be insufficient because you can be greater than a multiple of 3 (or a non-multiple of 3) with or without being a multiple of 3 yourself.
Insufficient.

Choose A.

sudhir18n · Answer

its MGMAT

The OE for statement 1.

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 ...

pike · Answer

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.

1 is sufficient.

fluke · Answer

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

LalaB · Answer

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

karjan07 · Answer

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

Bunuel · Answer

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.

Hope it's clear.

obs23 · Answer

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?

jlgdr · Answer

The product of three consecutive integers (n-1)(n)(n+1) must be divisible by 2 Statement 1 If n^2+n= n(n+1) is not divisible by three then (n-1) must be divisible by 3 Suff Statement 2 If k is a multiple of three Then we get that 3n >= k + 3 This only tell us that n >=1 but nothing else. Could be any number Hence, answer is A Hope it helps Cheers! J

That means that n

amitpassionate · Answer

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

Bunuel · Answer

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.

bumpbot · Answer

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

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