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k = (x – 1) (x + 2) – (x – 1) (x – 2)

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k = (x – 1) (x + 2) – (x – 1) (x – 2)

by sanju09 » Thu Feb 24, 2011 2:06 am
If x is an integer, and k = (x - 1) (x + 2) - (x - 1) (x - 2), then which of the following must be true?
(A) k is odd only when x is odd.
(B) k is odd only when x is even.
(C) k is even only when x is even.
(D) k is even only when x is odd.
(E) None of the above.

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by manpsingh87 » Thu Feb 24, 2011 2:19 am
sanju09 wrote:If x is an integer, and k = (x - 1) (x + 2) - (x - 1) (x - 2), then which of the following must be true?
(A) k is odd only when x is odd.
(B) k is odd only when x is even.
(C) k is even only when x is even.
(D) k is even only when x is odd.
(E) None of the above.

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IMO E

k will be even for both even and odd values of x..!!!
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by Anurag@Gurome » Thu Feb 24, 2011 3:35 am
k = (x - 1)[(x + 2) - (x - 2)] = 4(x - 1)
If x is ODD, then k = 4 * (Odd - 1) = 4 * Even = Even
If x is EVEN, then k = 4 * (Even - 1) = 4 * Odd = Even
So, whether x is even or odd, k will be even in both the cases.

Hence, correct answer is E.
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by mannesushanth » Thu Feb 24, 2011 9:50 am
What if x = 1,
if we substitute the equation 4 (x-1) with this value, then the result will be 0.

In this case how can we conclude that the result is always even?
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by Stuart@KaplanGMAT » Thu Feb 24, 2011 12:56 pm
mannesushanth wrote:What if x = 1,
if we substitute the equation 4 (x-1) with this value, then the result will be 0.

In this case how can we conclude that the result is always even?
0 is even!

0 is an "uncharged" (i.e. neither positive nor negative) even integer.
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by sbeagol1000 » Wed Mar 13, 2013 10:43 am
Everything distributes to 4(x-1)=k

Any number ODD or EVEN will make k an even number. Thus K will always be even.

(E) is the right choice.
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