Q170 DS GMAT Review 12th Edition - Questionable answer : DS Archive
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Q170 DS GMAT Review 12th Edition - Questionable answer

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11 Jun 2011, 16:00
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Dear All,

I just need someone to please take a look at this question in GMAT Review 12th Edition, DS, Q170.

If n is a positive integer, is n3 - n divisible by 4?

(1) n=2k + 1, where k is an integer,
(2) n2 + n is divisible by 6.

However.

n3 - n = n (n2 - 1) = n ( n+1) (n-1) = (n-1) n (n+1), i.e. 3 cons. integers.

n = 2K + 1, where k is an integer. n is a positive ODD integer, i.e. 2K + 1 > 0 and K> -1/2.

So, we can have cons. integers:

0, 1, 2 (where zero is an even integer)

2, 3, 4
and so on, where we are going to have at least one "four", therefore the answer seems to be sufficient EXCEPT we have this possibility of 0, 1, 2 where, as we know, zero is an even integer. n = 2K + 1 holds and so does (n-1) n (n+1). So, in my opinion, this is INSUFFICIENT.

Thank you.

Alex
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11 Jun 2011, 17:21
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11 Jun 2011, 18:46
Aleks1974 wrote:
Dear All,

I just need someone to please take a look at this question in GMAT Review 12th Edition, DS, Q170.

If n is a positive integer, is n3 - n divisible by 4?

(1) n=2k + 1, where k is an integer,
(2) n2 + n is divisible by 6.

However.

n3 - n = n (n2 - 1) = n ( n+1) (n-1) = (n-1) n (n+1), i.e. 3 cons. integers.

n = 2K + 1, where k is an integer. n is a positive ODD integer, i.e. 2K + 1 > 0 and K> -1/2.

So, we can have cons. integers:

0, 1, 2 (where zero is an even integer)

2, 3, 4
and so on, where we are going to have at least one "four", therefore the answer seems to be sufficient EXCEPT we have this possibility of 0, 1, 2 where, as we know, zero is an even integer. n = 2K + 1 holds and so does (n-1) n (n+1). So, in my opinion, this is INSUFFICIENT.

Zero is divisible by every positive integer. Remember we say that a is divisible by b if a/b is an integer. So zero is divisible by 4, for example, because 0/4 = 0, which is an integer. So the case where n^3 - n = 0 is not an exception in this question; in that case n^3 - n is certainly divisible by 4.
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12 Jun 2011, 03:10
The excellent solution is given in video solution. see it.
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12 Jun 2011, 07:21
IanStewart wrote:
Aleks1974 wrote:
Dear All,

I just need someone to please take a look at this question in GMAT Review 12th Edition, DS, Q170.

If n is a positive integer, is n3 - n divisible by 4?

(1) n=2k + 1, where k is an integer,
(2) n2 + n is divisible by 6.

However.

n3 - n = n (n2 - 1) = n ( n+1) (n-1) = (n-1) n (n+1), i.e. 3 cons. integers.

n = 2K + 1, where k is an integer. n is a positive ODD integer, i.e. 2K + 1 > 0 and K> -1/2.

So, we can have cons. integers:

0, 1, 2 (where zero is an even integer)

2, 3, 4
and so on, where we are going to have at least one "four", therefore the answer seems to be sufficient EXCEPT we have this possibility of 0, 1, 2 where, as we know, zero is an even integer. n = 2K + 1 holds and so does (n-1) n (n+1). So, in my opinion, this is INSUFFICIENT.

Zero is divisible by every positive integer. Remember we say that a is divisible by b if a/b is an integer. So zero is divisible by 4, for example, because 0/4 = 0, which is an integer. So the case where n^3 - n = 0 is not an exception in this question; in that case n^3 - n is certainly divisible by 4.

Thank you! That's what I needed.

Alex
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13 Jun 2011, 03:27
B is not sufficient because,
n = 2 , 3 give different values.

A is always divisible. Sufficient.
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Re: Q170 DS GMAT Review 12th Edition - Questionable answer   [#permalink] 13 Jun 2011, 03:27
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