arora007 wrote:If A , B , C , and D are integers such that A - C + B is even and D + B - A is odd, which of the following expressions is always odd?

a) A + D
b) B + D
c) C + D
d) A + B
e) A + C

let A=2,B=0,C=0,D=3
Satisfies both A-C+B is even and D + B - A is odd
eliminates options D and E

let A=1,B=1,c=0,D=3
eliminates A,B

pick C

kvcpk wrote: let A=2,B=0,C=0,D=3
Satisfies both A-C+B is even and D + B - A is odd
eliminates options D and E

let A=1,B=1,c=0,D=3
eliminates A,B

pick C

Great solution, that's really smart. This problem is really interesting.
Is there a way to solve this conceptually without plugging numbers? I know it's probably not ideal on the actual test but I'm just curious.
I listed all possibilities, such as:

A-C+B is even, so:

E-O=O
E-E+E
O-O+E
O-E+O

And D+B-A is odd:

O+O-E
O+E-E
E+E-O
E+O-E

But I have no idea how to go from here. Would love to see a way of solving this with pure logic or algebra.

conceptually yes.... infact...when i tried to solve this conceptually...i too got into a mess....and gave up...
this is how it is explained so neatly in gmatclub test..


Rewrite A - C + B as (A + B) - C and D + B - A as D + (B - A) . If A + B is even, B - A is even too. If A + B is odd, B - A is odd too. Now it follows from the stem that:

* if A + B is even, C is even and D is odd
* if A + B is odd, C is odd and D is even

Thus, C + D is always odd.

arora007 wrote:conceptually yes.... infact...when i tried to solve this conceptually...i too got into a mess....and gave up...
this is how it is explained so neatly in gmatclub test..


Rewrite A - C + B as (A + B) - C and D + B - A as D + (B - A) . If A + B is even, B - A is even too. If A + B is odd, B - A is odd too. Now it follows from the stem that:

* if A + B is even, C is even and D is odd
* if A + B is odd, C is odd and D is even

Thus, C + D is always odd.

Yep, that's what I was searching for and couldn't find, sigh.....Thanks.
I'm going to make sure I really integrate this until it's completely obvious to me.

Thanks for posting all these problem, btw. They are really hard and interesting, but still within the scope of GMAT math.

the concept which needs to sink in is perhaps...

if A+B is even A-B and B-A are also even
similarly
if A+B is odd A-B and B-A are also odd

arora007 wrote:the concept which needs to sink in is perhaps...

if A+B is even A-B and B-A are also even
similarly
if A+B is odd A-B and B-A are also odd

Exactly. The other thing I really want to take away from this is what to look for in this kind of problem.
I had all the necessary knowledge required to find that answer conceptually, I just didn't know which direction to look in and ended up wasting time listing every possibility, etc...

What I should've seen is that within each expression is included an A + or - B relationship. I think that's the most important thing to recognize. Finding a way to establish a relationship between both expressions.

Another approach:

A -C + B =even
D + B - A = odd

Thus, A -C + B + (D + B - A) = odd OR 2B - C + D is odd, since 2B is even, -C + D has to be odd

Another option, A -C + B - (D + B - A) = odd OR 2A - (C+D) is odd since 2A is even, C+D has to be odd

Hence pick C

mj78ind wrote:Another approach:

A -C + B =even
D + B - A = odd

Thus, A -C + B + (D + B - A) = odd OR 2B - C + D is odd, since 2B is even, -C + D has to be odd

Another option, A -C + B - (D + B - A) = odd OR 2A - (C+D) is odd since 2A is even, C+D has to be odd

Hence pick C

Brilliant, love it!

@Jeremy Thanks

mj78ind wrote:@Jeremy Thanks

You're welcome

This forum is great, I'm really learning a ton every day just from reading people's insights into different problems.
It's probably going to save me.

mj78ind wrote:Another approach:

A -C + B =even
D + B - A = odd

Thus, A -C + B + (D + B - A) = odd OR 2B - C + D is odd, since 2B is even, -C + D has to be odd

Another option, A -C + B - (D + B - A) = odd OR 2A - (C+D) is odd since 2A is even, C+D has to be odd

Hence pick C

I used the same approach..quite tricky question. But, sticking to eqn paid off..