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gmat club - tough one

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gmat club - tough one

by arora007 » Tue Jul 06, 2010 5:23 am
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
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by kvcpk » Tue Jul 06, 2010 5:44 am
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
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by jeremy8 » Tue Jul 06, 2010 5:52 am
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.
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by arora007 » Tue Jul 06, 2010 5:59 am
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.
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by jeremy8 » Tue Jul 06, 2010 6:05 am
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.
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by arora007 » Tue Jul 06, 2010 6:09 am
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
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by jeremy8 » Tue Jul 06, 2010 6:19 am
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.
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by mj78ind » Tue Jul 06, 2010 8:14 am
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
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by jeremy8 » Tue Jul 06, 2010 8:17 am
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!
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by mj78ind » Tue Jul 06, 2010 8:19 am
@Jeremy Thanks :)
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by jeremy8 » Tue Jul 06, 2010 9:41 am
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.
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by sumanr84 » Mon Jul 19, 2010 3:22 am
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..
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