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Is (a^b) (c^d) odd

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Is (a^b) (c^d) odd

by sanju09 » Thu Feb 16, 2012 2:36 am
Is (a^b) (c^d) odd, where a, b, c, and d are positive integers?
I. a is even and b is odd.
II. c is odd and d is even.




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Source: — Data Sufficiency |
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by rijul007 » Thu Feb 16, 2012 2:55 am
IMO: A

St(1):
if a is even
(a^b) (c^d) would be even, regardless of what c would be
suff

St(2):
c^d is odd
we dont have any idea of what a might be
insuff
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by Anurag@Gurome » Thu Feb 16, 2012 3:34 am
sanju09 wrote:Is (a^b) (c^d) odd, where a, b, c, and d are positive integers?
I. a is even and b is odd.
II. c is odd and d is even.
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(1) a is even and b is odd.
If a = 2, b = 1, then (a^b) * (c^d) = 2 * (c^d), which implies that a^b will always be even.
It is given that c and d are positive integers, so c^d will either be an odd or an even integer.
If c^d = even, then (a^b) * (c^d) = even * even = even
If c^d = odd, then (a^b) * (c^d) = even * odd = even
Therefore, (a^b) * (c^d) is always an even integer. So, the answer to the main question is "no"; SUFFICIENT.

(2) c is odd and d is even.
If c = 1 and d = 2, then c^d = 1, odd
So, c^d will always be odd.
It is given that a and b are positive integers, so a^b will either be an odd or an even integer.
If c^d = even, then (a^b) * (c^d) = even * odd = even
If c^d = odd, then (a^b) * (c^d) = odd * odd = odd
We do not get a definite answer; NOT sufficient.

The correct answer is A.
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by [email protected] » Fri Feb 17, 2012 4:36 am
Yes absolutely, the correct answer is A.

even X odd = even

Even X even = even

In any case as the statement 1 said that the first term is even, it is sufficient, no need for statement 2...
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by ArunangsuSahu » Fri Feb 17, 2012 11:37 am
(A)
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