Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."