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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 07 Oct 2007, 12:20
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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
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 [#permalink] New post 07 Oct 2007, 12:52
(n^3-n) divisible by 4 ?

statement 1

n = 2k + 1 ---> k is an integer

n is an odd integer.

Note that when n is an odd integer the outcome of (n^3-n) = n(n^2-1) will always divisible by 4 (i.e. 0,8,24)

sufficient

statement 2

n^2 + n is divisible by 6

insufficient

the answer is (A)

:)
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 [#permalink] New post 07 Oct 2007, 13:08
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

n^3 - n = n(n+1)(n-1) ie: the product of 3 consecutive

from 1

n is odd thus (n+1)(n-1) is even and divisible by 4. suff

from 2

n (n+1) is devisible by 6 this will give us multiple scenarios

n = 3 , n+1 = 4 , n-1 = 2............product is devisible by 4

n= 6 , n+1 = 7 , n-1 = 5 product is not devisible by 4

insuff

my answer is A
  [#permalink] 07 Oct 2007, 13:08
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If n is a positive integer, is n^3 - n divisible by 4? (1) n

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