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Number property - even/odd problem

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Number property - even/odd problem [#permalink] New post 23 Jul 2008, 23:38
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Can anyone help solving this problem?

Is A + B + C even?
(1) A - B - C is even?
(2) (A-C)/ is odd

OA will be coming soon!
By the way, is there any specific quick approach of dealing even-odd problem? Thanks!
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Re: Number property - even/odd problem [#permalink] New post 23 Jul 2008, 23:47
Is A + B + C even?
(1) A - B - C is even
(2) (A-C) is odd


I go with A

St1: A if odd then B+C is odd => A+B+C is even and A if even B+C is even => A+B+C is even. There can be no case where st 1 is true and we cant say what A+B+C would be.----suffic

St2: A-C is odd now nothing has been told abt B hence insuff
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Re: Number property - even/odd problem [#permalink] New post 24 Jul 2008, 00:13
I go with A as well

My approach :

1) A- b-c is even

means

A - ( B +C ) is even

Now even-even = even

or odd-odd = even

Therefore either ( A is even , then B + C is even ) or (if A is odd , then B + C is odd .. )

either ways

A + ( B + C ) = odd + odd = even

or if A is even than B+ C is also even hence

A+ ( B + C ) = odd + odd = even

Therefore A is correct.

Option A- C = odd, gives no information on B, hence inconclusive,
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Re: Number property - even/odd problem [#permalink] New post 24 Jul 2008, 01:48
For sum or difference:
Even and even gives even
Odd and odd gives even
Odd and even gives odd

As such

(1) is sufficient. If A-B is even, then A+B is even, vice versa. If (A-B)-C is even, (A+B)+C is even.

(2) I think you mean (A-C)/B is odd

(2) tells us two possible cases

Case (i) (A-C) is odd and B is odd, and Case (ii) (A-C) is even and B is even

In (i) if (A-C) is odd and B is odd

We can also conclude A or C is odd and the other is even

So two possible combination
(a) A (odd), B (odd), C (even)
(b) A (even), B (odd), C (odd)

In both (a) and (b) prove A + B + C is even

See Case (ii) (A-C) is even and B is even
So two possible combination
(a) A (odd), B (even), C (odd)
(b) A (even), B (even), C (even)

This also gives A+B+C even

So (2) alone is also sufficient.
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Re: Number property - even/odd problem [#permalink] New post 24 Jul 2008, 03:52
is something missing with statement 2 ? im not a fan of making assumptions :)
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Re: Number property - even/odd problem [#permalink] New post 24 Jul 2008, 07:11
It is a typical GMAT trap named unwarrant assumption. We are not given that A, B, C are integer!

OA is E!

Let's pick up numbers:

(1) Pick 9/2 - 3/2 - 1 for A - B - C respectively --> A - B - C = 2 (even) but A + B + C = 1 (Odd) --> insuff

(2) similarly, just pick up fraction like 2, 1/3, 1 for A, B,C, we got (A-C)/B odd but A+B+C is even.

(1)&(2), just the same approach.

Wondering if there is a better approach?
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Re: Number property - even/odd problem [#permalink] New post 24 Jul 2008, 07:28
That's a good question!
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Re: Number property - even/odd problem [#permalink] New post 24 Jul 2008, 09:36
There are a few key number properties you must know:

Adding two evens or adding two odds results in an even number
Adding an even and an odd results in an odd number

For more info, see this Number Properties page that I found.
Re: Number property - even/odd problem   [#permalink] 24 Jul 2008, 09:36
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