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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.
***THIS SOLUTION HAS FLAWS AS POINTED OUT BY TomB***
Sorry, didn't use the matrix for this one.
A-C+B=even; means Either one of these Or two of these Or All three of them ARE EVEN.
But, NOT all of them are ODD.
D + B - A=odd Means, ALL of them are ODD
That confirms; A, B, D are ODD; and from before, C must be EVEN.
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?
\(A + D\) \(B + D\) \(C + D\) \(A + B\) \(A + C\)
Start with all possible combinations for A and B:
ODD: O EVEN: E
C
A
B
D
O
E
E
O
O
O
E
E
Now fill in the column 1, i.e. C's type with either odd or even; based on statement 1
C
A
B
D
O
O
E
O
E
O
E
O
O
E
E
E
Now fill in the column 4, i.e. D's type with either odd or even; based on statement 2
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.
Looks good question....
We always need simple, easy and quick/fast approach:
A-C+B = EVEN ............I D+B-A = ODD..............II
Add I and II:
A-C+B+D+B-A = EVEN+ODD 2B-C+D = ODD
No matter whether B is even or odd, 2B has to be an even number. If so, then remaining number, (D-C), has to be 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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Re: If A, B, C, and D are integers such that A - C + B is even [#permalink]
27 Aug 2015, 18:29
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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