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Even - Odd = Odd Therefore: A-C+B - (D+B-A) = ODD .. 2A-D-C = ODD ( now we know anything multiplied by 2 makes it Even.. so 2A cannot be the factor to cause this to be odd.. ) so we can simplfy it : -(D+C) = ODD

If A , B , C , and D are integers such that A - C + B is even and D + B - A is odd, which of the following expressions is always odd?

A + D B + D C + D A + B A + C

I am wondering if there is any matrix method to do these sort of Even/Odd question. If anybody knows about any such method, kindly post.

Make it plain and simple:

A - C + B = even D + B - A = odd

Add both: A - C + B + D + B - A = Even +odd 2B - C + D = odd 2B + (D - C) = odd

Since 2B is even, (D-C) must be odd. Alternatively, C could be odd or even and same is D but if C is odd, D must be even and vice versa. However it is not necessary to find out whether C is odd or even.

If so, either (C+D) or (C-D) is odd. So its C that C + D is always odd.

Re: If A, B, C, and D are integers such that A - C + B is even [#permalink]
23 Jul 2014, 21:09

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