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odd/even - tough one

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odd/even - tough one [#permalink]

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New post 25 Jul 2009, 16:27
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A
B
C
D
E

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please solve
and also explain, if time permits
http://www.postimage.org/image.php?v=aV1UXnar
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Re: odd/even - tough one [#permalink]

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New post 25 Jul 2009, 16:45
IMO B.

1) wx + yz is odd.
So wx is even, yz is odd or wx is odd and yz is even.
if wx is even then w/x is odd or even (w = 6, x = 2 w/x = 3 or w=8, x=2 w/x = 4)
if wx is odd then w/x is odd
Similarly for yz also.
so w/x + y/z might be odd or even.
2) wz + xy is odd
w/x + y/z = (wz + yx)/xz and since w/x and y/z are integers wz/xz, yx/zx will also be same integers.
So (wz + yx)/xz is odd

Answer is B.

Please confirm if I am right or not
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Re: odd/even - tough one [#permalink]

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New post 25 Jul 2009, 18:57
OA B
and you are absolutely right
and if possibe can u elaborate stmt 1 a bit more
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Re: odd/even - tough one [#permalink]

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New post 25 Jul 2009, 20:42
When sum of two numbers is odd then one number should be even and other should be odd. Sum of two even numbers or sum of two odd numbers is always even.

wx + yz is odd implies either wx is even and yz is odd or vice versa.
Re: odd/even - tough one   [#permalink] 25 Jul 2009, 20:42
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