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a combination of calculation and intuition and substitution suggests to me it should be D. Please confirm.

looking at fresinha post i think i have a mathematical proof for D.

it can be inferred that a can be expressed as 4x and b as 2y. just to drive home the "must" point in the question, it is possible that b can also take a form of 4y, but we are taking the worst case scenario, hence a will ALWAYS have a factor of 4, not b.

(a+2)/2=(4x+2)/2=2x+1 --> always odd none of the other options satisfy with such certainty.

we also know that a=b*even which means a will always b divisible by 4..

so lets pick 8 +2=10/2=5 you cant pick something like 12 since it violates the stem..

basically a=2^n...

2^n +2=2(2^(n-1) +1)/2 then you get 2^(n-1) + 1 ..i.e even+odd=odd...always

Agree with D but the statement that "basically a=2^n..." is not true cuz "a" has to be a multiple of 4. in that case "a" could be 4 or 8 or 12. If a is 12, then it doesnot fit anywhere on the expression that a = 2^n.... so a = 2^n is not correct.

the strongest reason why (a+2)/2 is odd is that a must be a multiple of 4.

I dont think you can have any odd prime factors for a..

lets say for argument sake..a=12 and b=4 12-4=8 but 12/4=3 so you see you can have a multiple of 4, but it wont meet the requirement laid out in the stem..therefore i feel..a at the least has to be 2^n ...

GMAT TIGER wrote:

fresinha12 wrote:

i get d..

a+2/2 will be odd..

from the stem we know both a and b are even..

we also know that a=b*even which means a will always b divisible by 4..

so lets pick 8 +2=10/2=5 you cant pick something like 12 since it violates the stem..

basically a=2^n...

2^n +2=2(2^(n-1) +1)/2 then you get 2^(n-1) + 1 ..i.e even+odd=odd...always

Agree with D but the statement that "basically a=2^n..." is not true cuz "a" has to be a multiple of 4. in that case "a" could be 4 or 8 or 12. If a is 12, then it doesnot fit anywhere on the expression that a = 2^n.... so a = 2^n is not correct.

the strongest reason why (a+2)/2 is odd is that a must be a multiple of 4.

Yes you can. Lets take a = 24, which is a multiple of 4 and has 3 as odd prime factor, and b = 4. then a - b = 24-4 = 20 a/b = 24/4 = 6 (a+2)/2 = (24+2)/2 = 13

So it is not necessary for "a" to have its value a power of 2.

fresinha12 wrote:

I dont think you can have any odd prime factors for a..

lets say for argument sake..a=12 and b=4 12-4=8 but 12/4=3 so you see you can have a multiple of 4, but it wont meet the requirement laid out in the stem..therefore i feel..a at the least has to be 2^n ...

GMAT TIGER wrote:

fresinha12 wrote:

i get d..

a+2/2 will be odd..

from the stem we know both a and b are even..

we also know that a=b*even which means a will always b divisible by 4..

so lets pick 8 +2=10/2=5 you cant pick something like 12 since it violates the stem..

basically a=2^n...

2^n +2=2(2^(n-1) +1)/2 then you get 2^(n-1) + 1 ..i.e even+odd=odd...always

Agree with D but the statement that "basically a=2^n..." is not true cuz "a" has to be a multiple of 4. in that case "a" could be 4 or 8 or 12. If a is 12, then it doesnot fit anywhere on the expression that a = 2^n.... so a = 2^n is not correct.

the strongest reason why (a+2)/2 is odd is that a must be a multiple of 4.