If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer?

A. a/2
B. b/2
C. (a+b)/2
D. (a + 2)/2
E. (b+2)/2

Bunuel, you are an absolute whiz

Yep the catch is that a has to be a multiple of 4 for a/b to be even

OA is D

First question: Why \(a\) has to be multiple of 4?

As we concluded \(a\) and \(b\) have to be even integers to meet both conditions in the stem: \(a=2m\) and \(b=2n\).

Then we have \(\frac{a}{b}\) is an even integer: \(\frac{a}{b}=2k\) --> \(a=2bk=4kn\) --> \(a\) is a multiple of \(4\).

Second question: Option C also COULD give an even result, so why not C?

The question asks: "which of the following MUST be an odd integer?"

If we take \(a=16\) and \(b=4\), then \(\frac{a+b}{2}=10=even\). So option C may or may not be odd and we are asked to determine which option is ALWAYS odd, hence C is out.

In fact:
Options A is always even.
Options B can be even or odd.
Options C can be even or odd.
Options D is always odd.
Options E can be even or odd.

Only option which is always odd is D.

Hope it helps.
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Given that (a-b) = even

Thus, \(b(\frac{a}{b}-1)\) = even --> as \(\frac{a}{b}\) is even, \((\frac{a}{b}-1)\) will be odd. Thus, b*odd = even--> b must be even

Thus, as a = b*2p --> a = 2l*2p = 4lp , where l,p are non-zero positive integers.

Thus,\(\frac{a}{2}\) + 1 = 2lp+1, which will always be an odd number.

D.
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