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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
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03 Mar 2018, 09:10

Top Contributor

jamifahad wrote:

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. A + D B. B + D C. C + D D. A + B E. A + C

Given: A - C + B = EVEN and D + B - A = ODD Add them to get: (A - C + B) + (D + B - A) = EVEN + ODD Simplify: 2B + D - C = ODD Since 2B is ALWAYS even, we get: EVEN + D - C = ODD So: D - C = ODD - EVEN This means: D - C = ODD If D - C is ODD, then we know that one of the numbers is EVEN and the other number is ODD If one of the numbers is EVEN and the other number is ODD, then C + D must also be ODD