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if x <> 2, what is the value of ((x-2)^n)/5 +

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New post 17 Feb 2005, 14:37
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if x <> 2, what is the value of ((x-2)^n)/5 + ((2-x)^n)/5 ?

1) x^3=27

2) (1/2)n has a remainder of 1
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New post 17 Feb 2005, 14:51
is it 1/2n or n/2 in statement II ? Assuming it is n/2

"B"

State 1: x = 3
1/5 + (-1)^n/5.....insuff as we don't know what n is

State 2: n = 2k+1....so n is always odd.....(x-2)^n/5 + (-1)^n*(x-2)^n/5...as n is odd second exp is -ve of the first....so it adds up to 0.
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New post 17 Feb 2005, 18:13
If it is n/2, I agree with banerjeea_98 and it is B
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New post 17 Feb 2005, 19:35
This maybe an extremely stupid question but what does "<>" represent?
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New post 18 Feb 2005, 21:02
C it is?

1)

x^3=27, x=3

(3-2)^n/5 + (2-3)^n/5...as we can see if n is even then the (2-3)=1, the sum is 1/5+1/5=2/5, if n is off then the sum is 0.... in suff

2) we know that n is not even

n/2 is not an integer!

combine the two statements....sufficient...the sum is 0

C it is...
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New post 18 Feb 2005, 21:06
OOpps didnt realize x<>2!

B it is...we know that (x-2) or (2-x) will be + ve and -ve....all we need to find is if n is even or odd!

B it is...

damn...on exam day I am afraid such carelessness will cost me big time :((

any suggestions on avoiding such mistakes?
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New post 28 Feb 2005, 21:01
I agree with baneerjea on this one
B

OA please?
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New post 28 Feb 2005, 23:18
2) is a tricky way of saying n is odd. When n is odd,the given expression will be 0.
1) is not sufficient.
B.
  [#permalink] 28 Feb 2005, 23:18
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if x <> 2, what is the value of ((x-2)^n)/5 +

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