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# Number property - even/odd problem

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Intern
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Number property - even/odd problem [#permalink]

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24 Jul 2008, 00:38
1
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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]

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24 Jul 2008, 00:47
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]

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24 Jul 2008, 01:13
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]

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24 Jul 2008, 02:48
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]

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24 Jul 2008, 04:52
is something missing with statement 2 ? im not a fan of making assumptions
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Re: Number property - even/odd problem [#permalink]

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24 Jul 2008, 08:11
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]

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24 Jul 2008, 08:28
That's a good question!
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Re: Number property - even/odd problem [#permalink]

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24 Jul 2008, 10:36
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

Re: Number property - even/odd problem   [#permalink] 24 Jul 2008, 10:36
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