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# It eats up lots of time to think..this one.... any faster

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It eats up lots of time to think..this one.... any faster [#permalink]

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10 Dec 2006, 00:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

It eats up lots of time to think..this one.... any faster method...

If c and d are integers, is c even?

(1) c(d+1) is even.

(2)(c+2)(d+4) is even.

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10 Dec 2006, 01:06
answer is C. here is how i do it:
remember that for a product to be even - one multiplier must be even.
st1. if d is 1 c can be anything hence insufficient
st2. same as st1. if d is 2 c can be anything.

together:

if c is odd, then from st1 d+1 must even. from st2 d+4 must be even (c is odd means c+2 is odd as well).
but d+1 and d+4 cannot be both even at the same time.
hence c cannot be odd. so c is even.

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10 Dec 2006, 01:38
suppose that c is odd.then for c(d+1) to be even , d+1 must be even.

From st 2, d+4 must be odd because d+1 is even
hence for (c+2)(d+4) to be even ,c+2 must be even which means c is even.

Hence c must be even.

Choice C
Re: ANother Data sufficiency   [#permalink] 10 Dec 2006, 01:38
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