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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.
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, 07:25, edited 1 time in total.
Why can't z be 0, then answer is E [#permalink]
12 Apr 2006, 13:15
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.
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.
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.
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