Find all School-related info fast with the new School-Specific MBA Forum

It is currently 18 Sep 2014, 19:53

Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Senior Manager
Senior Manager
avatar
Joined: 22 Jun 2005
Posts: 364
Location: London
Followers: 1

Kudos [?]: 4 [0], given: 0

GMAT Tests User
If x, y, and z are integers and xy + z is an odd integer, is [#permalink] New post 12 Apr 2006, 06:35
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
Manager
Manager
User avatar
Joined: 09 Feb 2006
Posts: 129
Location: New York, NY
Followers: 1

Kudos [?]: 1 [0], given: 0

 [#permalink] New post 12 Apr 2006, 07:06
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
Director
avatar
Joined: 24 Oct 2005
Posts: 662
Location: London
Followers: 1

Kudos [?]: 6 [0], given: 0

GMAT Tests User
 [#permalink] New post 12 Apr 2006, 07:14
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, 07:25, edited 1 time in total.
Senior Manager
Senior Manager
avatar
Joined: 22 Nov 2005
Posts: 482
Followers: 1

Kudos [?]: 3 [0], given: 0

GMAT Tests User
 [#permalink] New post 12 Apr 2006, 07:45
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
Intern
avatar
Joined: 25 Feb 2006
Posts: 6
Followers: 0

Kudos [?]: 0 [0], given: 0

Why can't z be 0, then answer is E [#permalink] New post 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.

thanks
Manager
Manager
User avatar
Joined: 09 Feb 2006
Posts: 129
Location: New York, NY
Followers: 1

Kudos [?]: 1 [0], given: 0

 [#permalink] New post 12 Apr 2006, 14:04
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
Intern
avatar
Joined: 25 Feb 2006
Posts: 6
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 12 Apr 2006, 14:24
thanks for the detailed explanation
Senior Manager
Senior Manager
avatar
Joined: 05 Jan 2006
Posts: 383
Followers: 1

Kudos [?]: 17 [0], given: 0

GMAT Tests User
 [#permalink] New post 13 Apr 2006, 00:30
got A..though I hope there is more easy and systematic way of solving such equations!
Manager
Manager
User avatar
Joined: 09 Feb 2006
Posts: 129
Location: New York, NY
Followers: 1

Kudos [?]: 1 [0], given: 0

 [#permalink] New post 13 Apr 2006, 06:00
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
VP
avatar
Joined: 21 Sep 2003
Posts: 1071
Location: USA
Followers: 3

Kudos [?]: 31 [0], given: 0

GMAT Tests User
 [#permalink] New post 13 Apr 2006, 09:37
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.
_________________

"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."

- Bernard Edmonds

  [#permalink] 13 Apr 2006, 09:37
    Similar topics Author Replies Last post
Similar
Topics:
41 Experts publish their posts in the topic If x, y and z are integers and xy + z is an odd integer, is Aleehsgonji 8 28 Jul 2009, 21:25
1 if x,y,z are integers is x(y^2 + z^3) even? vr4indian 2 29 Sep 2008, 10:40
1 If x, y, and z are consecutive odd integers such that 31 sarzan 2 24 Aug 2008, 12:50
2 Experts publish their posts in the topic If x y, and z are integers and xy + z is an odd integer, is mbawaters 8 22 May 2008, 05:26
Are x, y, z odd integers? 1) x+y+z is odd 2) x*y*z is a withme 2 18 Jan 2007, 07:39
Display posts from previous: Sort by

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

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.