\(a-b\) even --> either both even or both odd

\(\frac{a}{b}\) even --> either both even or \(a\) is even and \(b\) is odd.

As both statements are true --> \(a\) and \(b\) must be even.

As \(\frac{a}{b}\) is an even integer --> \(a\) must be multiple of 4.


Options A is always even.

Options B can be even or odd.

Options C can be even or odd.

Options D: \(\frac{a+2}{2}=\frac{a}{2}+1\), as \(a\) is multiple of \(4\), \(\frac{a}{2}\) is even integer --> even+1=odd. Hence option D is always odd.

Options E can be even, odd.


Answer: D.
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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

I didnt get why a has to be multiple of 4

Now if u take a as 4 and b as 2 then a-b is even a/b is even and a+b/2 is odd which is option C.

Please explain where i am going wrong?

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:
a > 0, b > 0 and a, b are Integers

a - b and a/b are both even intergers
==> \(a - b = 2m_1\)
\(\frac{a}{b} = 2m_2\) ==> \(a = (b)(2m_2)\)
==> a & b are both even, otherwise the above conditions won't be satisfied.

Now lets go with options one by one ...
A) a/2
==> can be even or odd ( a = 6 or a = 8)

B) b/2
==> Can be either even or odd ( b = 6 or b = 8)

C) (a+b) /2
==> Since, \(a = (b)(2m_2)\), \(\frac{a + b}{2}\) =\(\frac {(b)(2m_2) + b}{2}\)
==> \(\frac{a + b}{2}\) =\(\frac {(b)(2m_2 + 1)}{2}\)
==> Can be even or odd

D) (a+2)/2
==> \(\frac{a + b}{2}\) =\(\frac{2bm_2 + 2}{2}\)
==> \(\frac{a + b}{2}\) =\(bm_2 + 1\)
==> Since, b is even, \(bm_2\) is even and \(bm_2 + 1\) is odd always

E) (b+2)/2
==> \(\frac{b+2}{2} = 1 + \frac{b}{2}\)
==> Can be even or odd

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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The video solution is explained here.

Answer: Option D


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GMATinsight thanks for explanation. However not sure why in Option D, you used 4x and not just 4 like in other options calculations? Am I missing something? Thanks

Hi Kimberly77

That's because 'a' is a multiple for 4

Logic:
a/b = even
also, a and b are both even

Even/Even will be even only when numerator is minimum multiple of 4 when denominator is a multiple of 2
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Thanks GMATinsight for your reply. Got it now. Couldn't get my head round with 4x at the time.
Result will still be odd for option D using your logic ( 4+2/2 =3). Thanks again

A good question that tests the understanding of Odd-Even properties. The best approach would be to apply the Odd-Even properties rather than plugin values

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?

Let's analyze the data given and list out the possible scenarios.

For a- b to be an even integer, there are only 2 possible scenarios for a and b.

Case 1: a is even and b is even . Even - Even = Even
Case 2: a is odd and b is odd. Odd - Odd = Even


It's also given that a/b is also even. That means a is a multiple of b and the possible cases that satisfy this condition are

Case 1: a is even and b is even
Even/Even can result in an even integer. For eg. 8/2 = 4 (Even)
Case 2: a is even and b is odd.
Even/Odd can also result in an even integer For eg. 6/3 = 2 ( Even)

Combining both conditions, we can conclude that both a and b has to be even to satisfy both statements.

Do you think this is true in all cases? Let us try it out. 8/2 = 4 (Even) but 6/2 = 3 (Odd)

That means this is not true for all cases. So, we need to be more specific about a.

a/b = Even

a = b* Even = Even * Even as we already found that b is even.

Hence, we can clearly say that 'a' has to be a multiple of 4 as it's a product of two even numbers. For b, we can only say that it's an even integer i.e a multiple of 2.

The question here is which of the following must be an odd integer?

Since it's a MUST BE type question, we will try to eliminate each answer choice by proving that it could be an even integer.

A. a/2 ==> Since a is a multiple of 4 , a/2 is always even. 8/2 = 4 ,12/2= 6 . Hence, eliminated.

B. b/2 ==> We know that b is an even number, so b/2 can be odd or even depending on the value of b i.e 2/2 = 1 but 4/2 = 2(Even) . Eliminated.

C. (a+b)/2 ==> (a+b ) will be an even integer and Even/2 can be odd or even as explained in option B. So we can eliminate Option C as well.

D. (a + 2)/2 ==> a/2 + 1. Since a is a multiple of 4, a/2 is always even. So, Even + 1 will always give an odd integer. Therefore, option D is the answer
E. (b+2)/2 ==> b + 2 should be an Even number and Even/2 can be odd or even . Eliminated.

Option D is the correct answer.

Thanks,
Clifin J Francis,
GMAT Mentor
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