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If n is a positive integer and r is the remainder when (n

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Director
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If n is a positive integer and r is the remainder when (n [#permalink] New post 23 Jan 2007, 06:47
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C
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E

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If n is a positive integer and r is the remainder when (n – 1)(n + 1) is divided by 24, what
is the value of r?
(1) 2 is not a factor of n.
(2) 3 is not a factor of n.
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 [#permalink] New post 23 Jan 2007, 07:08
(C) for me :)

24 = 3*4*2 = 3*2*2*2

Stat1
n is not divisible by 2 implies that n is odd.

and that,
o either (n-1) or (n+1) is divisable by 4
o both (n-1) and (n+1) is divisable by 2

(n+1)(n-1) is divisble by 8 and it could be divisable by 3 or not. In the first case, the remainder would be 0 once divided by 24. In the second case, the remainder would not be 0.

INSUFF.

Stat2
n is not divisible by 3 implies. Does not help.

o If n=1, (n+1)*(n-1) = 0 that is divisable by 24 and remainder = 0
o If n=2, (n+1)*(n-1) = 3*1 = 3 that is not divisable by 24 and remainder = 3

INSUFF.

Both (1) and (2)
We will have:
o either (n-1) or (n+1) is divisable by 4
o both (n-1) and (n+1) is divisable by 2
and
o either (n-1) or (n+1) is divisable by 3

4*2*3=24

SUFF.

n is in the set {1, 5, 7, 11, 13, 17...}
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 [#permalink] New post 01 Feb 2007, 09:36
I agree with C,

my approach

Stat 1 tells us that N is not even, since it is not a factor of 2

therefore (n-1)(n+1) are 2 consecutive even integers

but this could mean anything 2 and 4, 4 and 6 this will yield different remainders


Stat 2 tells us that 3 is not a factor of N, so therefore N could be anything

11 and 13, 17 and 19

all yield different remainders


Combining , we know that (n-1)(n+1) are consective even postive #'s

so could be (0,2) (2,4) (4,6) (6,8) (8,10)

taking statement 2 into account

SINCE out of any 3 consective integers, one MUST be a factor of 3 then either (n-1) or (n+1) is a factor of 3

picking numbers in my set we realize that all consecutive even numbers that contain a factor of 3, yields a 0 remainder for the number 24


C !!!
  [#permalink] 01 Feb 2007, 09:36
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