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M20-04 27 Mar 2025, 12:27
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Bunuel
Could you please help validate/ invalidate the below logic. I'm confused as to why the below logic maybe wrong (in case it is).


From my deductions, I believe there are two correct answers to this question:
i.e C+D and A+B

Here's why IMO A+B could also be a 'must be odd' expression given the constraints.

Since, A-C+B is even, there are two Odd/even combinations that could work
1) All 3 elements are Even (Even +/- Even +/- Even = Even)
2) Two elements are odd, one of them is even (Odd +/-Odd +- Even = Even)

Since D+B-A is odd, there are two odd/ even combinations that could work
1) All 3 elements are Odd (Odd +/- Odd +/- Odd = Odd)
2) Two elements are Even, one of them is Odd (Odd +/- Even +- Even = Odd)

Now, on considering both expressions, we can see that A and B are common elements in both. So, we can eliminate the combinations where all three elements are even or odd- because both A and B cannot be both odd and even at the same time.

We are left with combinations: E, E, O and O, O, E
both combinations have atleast 1 Even (E) and one Odd (O)
We can deduct from this that- Either A is even, B is odd or B is even, A is odd

Since they must always have an opposite parity (Odd +- Even = Odd)
A+B must be odd
Bunuel
Official Solution:

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 must be odd?

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


Subtract the first equation from the second:

\((D + B - A) - (A - C + B) = odd - even\);

\(C + D - 2A = odd\);

\(C + D = odd + 2A\);

\(C + D = odd + even\);

\(C + D = odd\).


Answer: C
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M20-04 23 Sep 2025, 02:11
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GyanviP
Bunuel
Could you please help validate/ invalidate the below logic. I'm confused as to why the below logic maybe wrong (in case it is).


From my deductions, I believe there are two correct answers to this question:
i.e C+D and A+B

Here's why IMO A+B could also be a 'must be odd' expression given the constraints.

Since, A-C+B is even, there are two Odd/even combinations that could work
1) All 3 elements are Even (Even +/- Even +/- Even = Even)
2) Two elements are odd, one of them is even (Odd +/-Odd +- Even = Even)

Since D+B-A is odd, there are two odd/ even combinations that could work
1) All 3 elements are Odd (Odd +/- Odd +/- Odd = Odd)
2) Two elements are Even, one of them is Odd (Odd +/- Even +- Even = Odd)

Now, on considering both expressions, we can see that A and B are common elements in both. So, we can eliminate the combinations where all three elements are even or odd- because both A and B cannot be both odd and even at the same time.

We are left with combinations: E, E, O and O, O, E
both combinations have atleast 1 Even (E) and one Odd (O)
We can deduct from this that- Either A is even, B is odd or B is even, A is odd

Since they must always have an opposite parity (Odd +- Even = Odd)
A+B must be odd

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A + B is not necessarily odd. Consider A is odd, B is odd, C is even (this makes A - C + B even) and D is odd (this makes D + B - A odd).
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M20-04 23 Sep 2025, 06:48
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Thanks a lot for your prompt revert Bunuel ! That makes sense, seems like I missed that case while solving it the way I did.
Bunuel


A + B is not necessarily odd. Consider A is odd, B is odd, C is even (this makes A - C + B even) and D is odd (this makes D + B - A odd).
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