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OMG! Very Tricky!

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by Ozlemg » Sun Jul 24, 2011 7:17 am
If m is an integer, is m odd?

(1) m/2 is NOT an even integer
(2) m-3 is an even integer

OA to follow...
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Source: — Data Sufficiency |
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by GAMATO » Sun Jul 24, 2011 8:20 am
IMO B

(1) m/2 is NOT an even integer
So m/2 can be 0 or odd.
Case 1: m/2 = 0
m = 0
Case 2: m/2 is odd
m/2 = 2p+1
m= 2(2p+1) So m is even

As m can be either 0 or even, A is not sufficient

(2) m-3 is an even integer
m can not be zero as 0-3 = -3 (odd)
m-3 = 2p
m= 2p+ 3
So m is odd. B is Sufficient
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by Frankenstein » Sun Jul 24, 2011 8:27 am
GAMATO wrote:IMO B

(1) m/2 is NOT an even integer
So m/2 can be 0 or odd.
Case 1: m/2 = 0
m = 0
Case 2: m/2 is odd
m/2 = 2p+1
m= 2(2p+1) So m is even

As m can be either 0 or even, A is not sufficient

(2) m-3 is an even integer
m can not be zero as 0-3 = -3 (odd)
m-3 = 2p
m= 2p+ 3
So m is odd. B is Sufficient
Hi,
zero is even.
Well, coming to the question:
From(1): m/2 is not even integer. (m/2) need not be an integer. So, It can be either odd or an (odd integer/2)
So, m is either even or odd
Not sufficient
From(2):
m-3 is even. So, m is odd
Sufficient

Hence, B
Cheers!

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by Brent@GMATPrepNow » Sun Jul 24, 2011 9:02 am
Ozlemg wrote:If m is an integer, is m odd?

(1) m/2 is NOT an even integer
(2) m-3 is an even integer

OA to follow...
Statement 1:
If m/2 is NOT an even integer, we can come up with 2 contradicting values for m.

case a) m=3 (3/2 is not an even integer), which means m is odd
case b) m=6 (6/2 is not an even integer), which means m is even
So, statement 1 is not sufficient

Statement 2:
We know that: (odd) plus/minus (odd) = even
So, if m - 3 = even, we can be certain that m is odd
So, statement 2 is sufficient, and the answer is B

Cheers,
Brent
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by GAMATO » Sun Jul 24, 2011 10:07 am
Hey Frankenstein,

Thanks for the explanation. I did not realize that zero is an even integer. Thanks
Frankenstein wrote:
GAMATO wrote:IMO B

(1) m/2 is NOT an even integer
So m/2 can be 0 or odd.
Case 1: m/2 = 0
m = 0
Case 2: m/2 is odd
m/2 = 2p+1
m= 2(2p+1) So m is even

As m can be either 0 or even, A is not sufficient

(2) m-3 is an even integer
m can not be zero as 0-3 = -3 (odd)
m-3 = 2p
m= 2p+ 3
So m is odd. B is Sufficient
Hi,
zero is even.
Well, coming to the question:
From(1): m/2 is not even integer. (m/2) need not be an integer. So, It can be either odd or an (odd integer/2)
So, m is either even or odd
Not sufficient
From(2):
m-3 is even. So, m is odd
Sufficient

Hence, B
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