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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. \(A + D\) B. \(B + D\) C. \(C + D\) D. \(A + B\) E. \(A + C\)

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. \(A + C\)

Rewrite \(A - C + B\) as \((A + B) - C\) and \(D + B - A\) as \(D + (B - A)\). If \(A + B\) is even, \(B - A\) is even too. If \(A + B\) is odd, \(B - A\) is odd too. Now it follows from the stem that:

if \(A + B\) is even, \(C\) is even and \(D\) is odd

if \(A + B\) is odd, \(C\) is odd and \(D\) is even

bartone89 I believe your method is wrong because it assumes the values of the two equations are consecutive. ie 2k and 2k+1. I hope I havent misunderstood you solution.

Anyway, I think I may have found a better method.

The two equations are A-C+B and D+B-A, even and odd respectively.

On subtracting the two, you get, A-C+B-D-B+A = 2A-C-D (which is odd, since even - odd = odd)

2A-(C+D) is odd. Which means C+D is odd since we know that 2A has to be even.

I solved it in a different way: add both expressions A-C+B+D+B-A=odd. ok, rewrite everything as: 2B+D-C=odd 2B is always even, it must bean only 1 thing: D-C is always odd. if D-C is odd, then D+C is always odd.

I solved it in a different way: add both expressions A-C+B+D+B-A=odd. ok, rewrite everything as: 2B+D-C=odd 2B is always even, it must bean only 1 thing: D-C is always odd. if D-C is odd, then D+C is always odd.

I solved it in a different way: add both expressions A-C+B+D+B-A=odd. ok, rewrite everything as: 2B+D-C=odd 2B is always even, it must bean only 1 thing: D-C is always odd. if D-C is odd, then D+C is always odd.

Very good solution mvictor!

I agree very good solution. But it still requires timing luck or incredible number sense. I solved it through plugging in values. My method take no number sense needed and does not require too much logic.

A-C+B is even. Just plug in some easy numbers so that is true. A = 1 C = 2 B = 3 means that 1-2+3 = 2 which is even. D+B-A is odd. Re-use the same variables from before. D + 3 - 1 is D + 2. Set D to an easy odd number which is 5. So D = 5. Now you have values for A through D or (A,B,C,D) = (1,3,2,5). Work through the answers.

A. A+D = 1 + 5 = 6 B. B+D = 3 + 5 = 8 C. C+D = 2 + 5 = 7. This is odd. No need to do the rest, the answer is C. D. A+B E. A+C

I solved it in a different way: add both expressions A-C+B+D+B-A=odd. ok, rewrite everything as: 2B+D-C=odd 2B is always even, it must bean only 1 thing: D-C is always odd. if D-C is odd, then D+C is always odd.

Very good solution mvictor!

I agree very good solution. But it still requires timing luck or incredible number sense. I solved it through plugging in values. My method take no number sense needed and does not require too much logic.

A-C+B is even. Just plug in some easy numbers so that is true. A = 1 C = 2 B = 3 means that 1-2+3 = 2 which is even. D+B-A is odd. Re-use the same variables from before. D + 3 - 1 is D + 2. Set D to an easy odd number which is 5. So D = 5. Now you have values for A through D or (A,B,C,D) = (1,3,2,5). Work through the answers.

A. A+D = 1 + 5 = 6 B. B+D = 3 + 5 = 8 C. C+D = 2 + 5 = 7. This is odd. No need to do the rest, the answer is C. D. A+B E. A+C

the thing is - when presented with such expressions, you have to either add or subtract one from another. the thing is, you have 3 variables in each expression, therefore, you might lose some extra time to test all the possible options... my approach is very direct and fast though...and I always have the same approach to solving similar questions.