odd integer

[TingLi]

I think shobby has given the right solutions.
I don't think there is a exact formula for this question, what you should know is a is even, a/2 is even... that's enough, and nothing more.

[Fiver]

ah well.....any better method!!!????????????????

Given a-b is even=> both even or both odd
Given a/b is even=> both even or a is even and b is odd. Condition satisfied by both the statements is that both a and b are even. Now, if both a & b are even and a/b is an integer, then a must be greater than b and a multiple of 4.
Now option D reads: (a+2)/2= a/2+1 and any multiple of 4 when halfed will be an even number and 1+even is odd. Hence D is the correct answer

[subsvirgin]

a-b is even

This means that either both a and b are even or both a and b are odd.
--->Conclusion 1

a/b is even

This plus Conclusion 1 derives the fact that a and b both should be even.

Now if both are even, more than one of the options is right,.... can anybody pull me further???

[subsvirgin]

a-b is even

This means that either both a and b are even or both a and b are odd.
--->Conclusion 1

a/b is even

This plus Conclusion 1 derives the fact that a and b both should be even.

Now if both are even, more than one of the options is right,.... can anybody pull me further???

[subsvirgin]

a-b is even

This means that either both a and b are even or both a and b are odd.
--->Conclusion 1

a/b is even

This plus Conclusion 1 derives the fact that a and b both should be even.

Now if both are even, more than one of the options is right,.... can anybody pull me further???

[rkwagle]

Its D.

[bose]

ans is d
a/b =2m so a is even
a-b is even so b is even
so a=2mb=4mn as b=2n
so (a+2)/2=2mn+1 so odd
but (b+2)/2 = n+1 n can be even or odd so b can be odd or even

[kool_sunny]

But why not E.
Its also almost true......

[ramzrambo]

subsvirgin...check out my explanation that explains the special case for A, B......b can be 2(b/2 + 1 = 2...hence even) but "a" can never be 2.

[lesfeaves]

Fiver