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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.

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.

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.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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.