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

is the question correct?? none of the options seem to satisfy

here a>b since a/b is even integer

let a= 24 and b = 4.

A. a/2 = 12 => even
B. b/2 = 2 => even
C. (a+b)/2 = 14 => even
D. (a+2)/2 = 13 => odd
E. (b+2)/2 = 3 =>odd

now if a = 20 and b = 2 then a-b = 18 and a/b = 10
A. a/2 = 10 => even
B. b/2 = 1 => odd
C. (a+b)/2 = 11 => odd
D. (a+2)/2 = 13 => odd
E. (b+2)/2 = 2 => even

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

its a "MUST be" question. (a+b)/2 will not always be ODD.

also a-b and a/b is even and a and b will be even satisfying this condition. Since a/b is even so a will be multiple of 4.

Any set of values satisfying this will have a as 4n [n = natural number]

@Bunuel what a explanation man!!!!

Thanks a ton!!!!

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
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a and b both have to be even to fulfill the condition.
a could be = 4, 6, 8 ...., b = 2,4, ....
so a - b = 2 (Even)
a/b= 2k (Even)
a/2=2k
a = 4k
so a must be multiple of 4
now try the options
ans. D
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This is where I got confused.

E. (b+2)/2

Since we have established that b is even, I was initially wondering how can (b+2)/2 be even.

The catch is that b can be 2 also, in which case, (b+2)/2 will be 2 (even).

Best approach would be taking number
as a-b = even, this means either both are odd, or both are even...(1)
a/b = even, hence a = b*even = even. ...(2)

from (1) and (2), both a and b are even, also a = even *even = multiple of 4

Now take a = any multiple of 4, say 4

D = (4+2)/2 = 3

If we had taken a = 8,
D =(8+2)/2 = 5

It will always be odd.

(PS: (D) = (4K+2)/2 =2K+1 , HENCE ALWAYS ODD)



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