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If n is a positive integer, is n^3-n divisible by 4? 1) n =

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If n is a positive integer, is n^3-n divisible by 4? 1) n = [#permalink] New post 15 Apr 2006, 20:07
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If n is a positive integer, is n^3-n divisible by 4?

1) n = 2k+1, where k is an integer

2) n^2+n is divisible by 6

Pls explain your answer.
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 [#permalink] New post 15 Apr 2006, 20:13
IMO A.

n^3 - n = n(n^2 - 1) = (n-1)n(n+1)

which is divisible if n is odd.

1) gives that n is odd. sufficient to answer. Sufficient.

2) n can be even (n=6) or odd (n=5). Not sufficient
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 [#permalink] New post 15 Apr 2006, 22:10
A for me

1) (2k+1)^3 - (2k-1)
= 8k^3-2k+2/4
Can cancel the 4 out

2) n=2 no, not integer
n=5 yes, integer
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 [#permalink] New post 16 Apr 2006, 00:15
Sorry, i cannot seem to get the concept behind solving this and trying to break it down. Hope you can be patient enough to explain...

vipin7um, why is the simplified question stem [n(n+1)(n-1) all over 4] only divisible if n is odd? whether n is odd or even, the numerator will always give an even number. then i only need to determine from the statements whether that number is divisible by 4 right?
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 [#permalink] New post 16 Apr 2006, 16:19
Hermione wrote:
Sorry, i cannot seem to get the concept behind solving this and trying to break it down. Hope you can be patient enough to explain...

vipin7um, why is the simplified question stem [n(n+1)(n-1) all over 4] only divisible if n is odd? whether n is odd or even, the numerator will always give an even number. then i only need to determine from the statements whether that number is divisible by 4 right?


Actually, let me correct myself. n(n+1)(n-1) would be definitely divisible by 4 if n is odd. It will be divisible by 4, if n is multiple of 4 as well.

This is so because the above expression is nothing but product of three consecutive numbers, n being the middle number. So if n is odd, then the number that precedes it, and the number that follows it will be even. Which means the product will have at least two even numbers and hence it will be divisible by four.

HTH...
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 [#permalink] New post 20 Apr 2006, 05:23
Actually, let me correct myself. n(n+1)(n-1) would be definitely divisible by 4 if n is odd. It will be divisible by 4, if n is multiple of 4 as well.


but what if we have the sequence: 0, 1, 2

Is 0 a positive integer? And 0 is an even integer?

Last edited by macca on 20 Apr 2006, 05:33, edited 1 time in total.
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 [#permalink] New post 20 Apr 2006, 05:30
macca wrote:
Is 0 a positive integer?


0 is neither positive nor negative. It is however an even integer.
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 [#permalink] New post 20 Apr 2006, 06:16
OA is A.

If n is a positive integer, is n^3-n divisible by 4?

n^3-n = n*(n^2-1) = (n-1) * n * (n+1)

1) n = 2k+1, where k is an integer

=> 2k must be even in all cases of k, which makes 2k+1 odd. In other words, n is odd. Therefore:

(n-1) = even
n = odd
(n+1) = even

since there are two evens in the expression, it must be divisible by 4. SUFFICIENT

2) n^2+n is divisible by 6

n^2+n = n*(n-1)

=> Two cases possible:

1.
n = odd
(n-1) = even

OR

2.
n = even
(n-1) = odd

Regardless of which, this statement only tells us that there is one even (=divisible by 2). Note: 6 = 3 * 2. INSUFFICIENT
  [#permalink] 20 Apr 2006, 06:16
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