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m13 #13 The unlucky 13!

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Senior Manager
Joined: 18 Aug 2009
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m13 #13 The unlucky 13! [#permalink]  16 Nov 2009, 21:19
Is $$(x - y)*(x + y)$$ an even integer?

1. $$x$$ and $$y$$ are integers
2. $$x + y$$ is even

Chose incorrect B and after looking at the solution thought of adding a little to OE (don't know if it is a superstition, but I'm getting #13 wrong in so many tests )...

x+y is even

in case both are integers, then they are either both odd or both even => x-y will be even.

in case both are fraction (if only 1 is fraction, then "x+y is even" does not holds true). Check it with few substitutions for odd/even combination on denominator so as to keep x+y even:
Even denominator... 1/2+3/2 => subtraction result odd. But overall result even.
Odd denominator... 5/3+1/3 => subtraction result is fraction making B false
So they have to be integers OA
[Reveal] Spoiler:
C
Senior Manager
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Re: m13 #13 The unlucky 13! [#permalink]  16 Nov 2009, 21:23
'am I glad that it not my 13th post on Friday the 13th
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Re: m13 #13 The unlucky 13! [#permalink]  22 Nov 2009, 07:48
S1 is insufficient
s2 is sufficient because
(x-y) is either even or odd integer, but multiplication with even (x+y) will always be even. So it is sufficient to tell us the given statement result.

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Re: m13 #13 The unlucky 13! [#permalink]  22 Nov 2009, 08:42
gbansal745 wrote:
S1 is insufficient
s2 is sufficient because
(x-y) is either even or odd integer, but multiplication with even (x+y) will always be even. So it is sufficient to tell us the given statement result.

Ans. shud be C as we need to know that x,y are integers.

Consider x=13/4, y = 11/4, then x+y=6 and x-y=1/2 and (x-y) (x+y) = 3 (odd)
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Re: m13 #13 The unlucky 13!   [#permalink] 22 Nov 2009, 08:42
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m13 #13 The unlucky 13!

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