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If x, y, and z are integers and xy + z is an odd integer, is [#permalink]
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12 Apr 2006, 07:35
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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



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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
The only way to get an odd integer from adding two integers is if one is odd and one is even. From the question stem then, we know that either xy is odd and z is even, or that xy is even and z is odd.
Statement 1:
In this situation, we can only get an even integer if BOTH integers are even or BOTH integers are odd. This statement contains one similar integer from the stem  xy. If xy were odd, then z would be even  if this were the case, the sume of xy and xz would be odd, because xy would be odd and xz would be even (even times an odd always results in an even). Because this is not the case, we know that xy must be even in the original statement, and z must be odd; therefore x is even. Sufficient.
Statement 2:
Here we cannot tell. If xy were even in the stem, and z odd, x and/or y could be even. If x were odd, and z were odd, xz would be odd, and y would be even. This results in an odd integer for their sum. If z had been even in the original stem, and xy odd, then x*z would be even, and y odd  resulting in an odd sum. Therefore, statement 2 is insufficient.
Answer is A.
Hope the "rambling" made sense.



Director
Joined: 24 Oct 2005
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Hmm..
1) xy + xz = even
ie. X Even or Y + Z Even
ie. X even Y odd Z odd
or X even Y even Z even
But we know, xy + z = odd
hence, XY = odd and Z even or XY even and Z odd
X odd Y odd Z even
X odd Y even Z odd
X even Y odd Z odd
X even Y even Z odd
the common statement satisfying both is X even Y odd Z odd
Taking values, for both cases(odd = 1, even =2), we can see that this statement is sufficient
2) y + xz = odd
ie. Y odd and XZ even OR Y even and XZ odd
ie. Y odd X even Z even
Y odd X even Z odd
Y odd X odd Z even
Y even X odd Z odd
But we know, xy + z = odd
hence, XY = odd and Z even or XY even and Z odd
X odd Y odd Z even
X odd Y even Z odd
X even Y odd Z odd
X even Y even Z odd
But there are few common statement satisfying both Taking values, for both cases(odd = 1, even =2), we can see that this statement is sufficient
hence A
Last edited by remgeo on 12 Apr 2006, 08:25, edited 1 time in total.



Senior Manager
Joined: 22 Nov 2005
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Nice explanation by jcool child.
I'll also go for A.
But can we come with some easy cool approach.
To avoid ambiguity I used table method.
Professor, HongHu can you guys comment.



Intern
Joined: 25 Feb 2006
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Why can't z be 0, then answer is E [#permalink]
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12 Apr 2006, 14:15
Please note the question does not say positive integer or negative so for example y can be 0, in that case the information provided makes it E. Where am i wrong.
thanks



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Swaroh  It would not matter. We can run through again using your point (what if y is zero) as an assumption.
First, we would therefore KNOW that z is an integer that is not zero, because the stem would then not be true; we also know then z MUST BE ODD.
So statement 1:
xy + xz is even.
If y = 0, then xz must be even, which would tell us that x is even, because z would have to be odd from the stem.
Also, if x = 0, then the statement would still hold true because zero is even (depending on how you look at it  it could be neither even nor odd).
Statement 2 would still be insufficient. Well, not 100% true, but since we can't say for certain whether y is 0, we wouldn't be able to test this. If x were 0, and y were odd, then this would hold; if y were 0, and x and z were both odd, this would also hold.
So we'd still get A, accounting for y or x = 0.
I may not have accounted for every single possibility here, but at least this should show that it wouldn't matter if y were 0, or x were 0.
You could also consider the case where z = 0. Statement 1 would therefore be at odds with the question stem; so this could not happen.



Intern
Joined: 25 Feb 2006
Posts: 6

thanks for the detailed explanation



Senior Manager
Joined: 05 Jan 2006
Posts: 381

got A..though I hope there is more easy and systematic way of solving such equations!



Manager
Joined: 09 Feb 2006
Posts: 129
Location: New York, NY

Well, it takes longer to explain that it does to actually think through the process. You can also just use a table, assigning odd or even to each variable or product of variables to see what happens.



VP
Joined: 21 Sep 2003
Posts: 1057
Location: USA

chiragr wrote: got A..though I hope there is more easy and systematic way of solving such equations!
chiragr,
JC has a really good explanation.
This was discussed a couple of weeks back and I provided a soln without picking nos. Here is the link:
http://www.gmatclub.com/phpbb/viewtopic.php?t=27974
HTH.
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