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tricky odd/even integer question

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tricky odd/even integer question

by Baldini » Sun May 17, 2009 11:02 pm

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If x, y and z are integers and xy + z is an odd integer, is x an even integer?

1. xy + xz is an even integer
2. y + xz is an odd integer

Could someone please explain how to get to the answer as quickly as possible?

Thanks
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by bluementor » Mon May 18, 2009 7:26 am

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There are 4 cases that will satisfy xy + z = odd
(i) x = even, y = even, z = odd
(ii) x = even, y = odd, z = odd
(iii) x = odd, y = even, z = odd
(iv) x = odd, y = odd, z = even

is x even?

Statement 1:

xy + xz = even
x(y + z) = even

case (a): x = odd, (y + z) = even. this is not possible since this will not satisfy any of the cases(i) - (iv).

case (b): x = even, (y + z) = odd/even. this satisfies cases (i) and (ii).

so, since only case b is valid, x must be even. sufficient.

Statement 2:

y + xz = odd

case (c): y = odd, xz = even. this satisfies cases (ii) and (iv) where x could either odd or even. insufficient.
case (d): y = even, xz = odd. this satisfies case (iii), where x is odd.

so, since this statement doesn't provide any restriction on x, it is not sufficient.

Choose A.

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by Baldini » Mon May 18, 2009 8:24 am

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Thanks a lot Bluementor.
Can I just ask how long it took you to get the OA? Is there anything you did to streamline your process (took me a long time to do this)?
thanks
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Re: tricky odd/even integer question

by Ian Stewart » Mon May 18, 2009 11:04 am

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Baldini wrote:If x, y and z are integers and xy + z is an odd integer, is x an even integer?

1. xy + xz is an even integer
2. y + xz is an odd integer

Could someone please explain how to get to the answer as quickly as possible?

Thanks
At least when considering S1, you can proceed quickly if you notice the similarity between the expression in S1 and the expression in the question. We know xy + xz is even, and that xy + z is odd; we can subtract to eliminate the xy:

xy + xz - (xy + z) = even - odd
xz - z = odd
z(x - 1) = odd

so z is odd, and x - 1 is odd, so x is even. S1 is sufficient.
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Re: tricky odd/even integer question

by iamcste » Mon May 18, 2009 11:15 am

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Ian Stewart wrote:
Baldini wrote:If x, y and z are integers and xy + z is an odd integer, is x an even integer?

1. xy + xz is an even integer
2. y + xz is an odd integer

Could someone please explain how to get to the answer as quickly as possible?

Thanks
At least when considering S1, you can proceed quickly if you notice the similarity between the expression in S1 and the expression in the question. We know xy + xz is even, and that xy + z is odd; we can subtract to eliminate the xy:

xy + xz - (xy + z) = even - odd
xz - z = odd
z(x - 1) = odd

so z is odd, and x - 1 is odd, so x is even. S1 is sufficient.
Cool solution
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by cubicle_bound_misfit » Mon May 18, 2009 11:20 am

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if this is of any help

from stmt 1,

xy +z +(x-1)z == even

odd + (x-1)z == even

hence (x-1)z is odd

hence x-1 odd and z odd

hence x even and z odd

putting back in the original condition

xy = even
z =odd

which satisfies the condition hence yes x is even, choose A/D

for stmt 2,

my reasoning is same as 'Bluementor'

hence answer A.

regards,
CBM
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by bblast » Wed Aug 24, 2011 9:20 am

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Ian Stewart wrote:
Baldini wrote:If x, y and z are integers and xy + z is an odd integer, is x an even integer?

1. xy + xz is an even integer
2. y + xz is an odd integer

Could someone please explain how to get to the answer as quickly as possible?

Thanks
At least when considering S1, you can proceed quickly if you notice the similarity between the expression in S1 and the expression in the question. We know xy + xz is even, and that xy + z is odd; we can subtract to eliminate the xy:

xy + xz - (xy + z) = even - odd
xz - z = odd
z(x - 1) = odd

so z is odd, and x - 1 is odd, so x is even. S1 is sufficient.
Hi Ian,
I remember you posted a similar solution to a similar GMAT club problem I posted here. Thanks for this approach on statement 1. Can we use similar method to deduce a result from statement 2 ? Or is testing cases is the only method out ?
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by navami » Thu Aug 25, 2011 4:36 am

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Good question thanks. considering option 2. was little difficult.
This time no looking back!!!
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by czarczar » Fri Aug 26, 2011 5:36 am

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Ian Stewart wrote:
Baldini wrote:If x, y and z are integers and xy + z is an odd integer, is x an even integer?

1. xy + xz is an even integer
2. y + xz is an odd integer

Could someone please explain how to get to the answer as quickly as possible?

Thanks
At least when considering S1, you can proceed quickly if you notice the similarity between the expression in S1 and the expression in the question. We know xy + xz is even, and that xy + z is odd; we can subtract to eliminate the xy:

xy + xz - (xy + z) = even - odd
xz - z = odd
z(x - 1) = odd

so z is odd, and x - 1 is odd, so x is even. S1 is sufficient.
Perfect approach!
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by prateek_guy2004 » Fri Aug 26, 2011 7:50 am

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Statement 1 sufficient


1. xy + xz is an even integer

x(y+z) = even

2. y + xz is an odd integer Statement 2 Not sufficient.

Answer A
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by GMATGuruNY » Fri Jul 13, 2018 3:37 am

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Baldini wrote:If x, y and z are integers and xy + z is an odd integer, is x an even integer?

1. xy + xz is an even integer
2. y + xz is an odd integer
Statement 1:
Test whether it's possible that x is odd.
Let x=1.
Substituting x=1 into xy+z = ODD, we get:
y+z = ODD
Substituting x=1 into xy+xz = EVEN, we get:
y+z = EVEN
The equations in red contradict each other, implying that it is not possible that x is odd.
Thus, x must be even.
SUFFICIENT.

Statement 2:
Test whether it's possible that x is odd.
Let x=1.
Substituting x=1 into xy+z = ODD, we get:
y+z = ODD
Substituting x=1 into y+xz = ODD, we get:
y+z = ODD
Since the equations in green are the same, it's possible that x is odd.

Test whether it's possible that x is even.
Let x=2.
Substituting x=2 into xy+z = ODD, we get:
2y+z = ODD
Substituting x=2 into y+xz = ODD, we get:
y+2z = ODD
The equations in green are both viable if y=1 and z=1, implying that it's possible that x is even.

Since the answer to the question stem is NO in the first case but YES in the second case, INSUFFICIENT.

The correct answer is A.
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