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Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink]
 11 Jun 2011, 17:00
Question Stats:
50% (02:40) correct
50% (00:00) wrong based on 0 sessions
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. GMAT answer is A. 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. Please help. Thank you. Alex
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Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink]
11 Jun 2011, 18:21
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Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink]
11 Jun 2011, 19:46
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.
GMAT answer is A.
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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Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink]
12 Jun 2011, 04:10
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Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink]
12 Jun 2011, 08:21
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.
GMAT answer is A.
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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Re: Q170 DS GMAT Review 12th Edition - Questionable answer [#permalink]
13 Jun 2011, 04:27
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, 04:27
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