Last visit was: 19 Nov 2025, 17:51 It is currently 19 Nov 2025, 17:51
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.
Close
Request Expert Reply
Confirm Cancel
User avatar
MA
Joined: 25 Nov 2004
Last visit: 09 Aug 2011
Posts: 697
Own Kudos:
Posts: 697
Kudos: 515
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
HongHu
Joined: 03 Jan 2005
Last visit: 25 Apr 2011
Posts: 966
Own Kudos:
Posts: 966
Kudos: 796
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
mdf2
Joined: 11 Jan 2005
Last visit: 28 Jul 2005
Posts: 48
Own Kudos:
Posts: 48
Kudos: 3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
ssumitsh
Joined: 06 Sep 2004
Last visit: 11 Feb 2005
Posts: 26
Own Kudos:
Posts: 26
Kudos: 1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
(i) (z1+z2+z3+ ..................+zn)/n is an odd integer.
(ii) n is an odd integer

I will go with A.

(i) is sufficient to prove that the sum of consecutive integers z1 to zn is odd.
avatar
HongHu
Joined: 03 Jan 2005
Last visit: 25 Apr 2011
Posts: 966
Own Kudos:
Posts: 966
Kudos: 796
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Oh my, I just realized.

S=n(n+1)/2
S/n=(n+1)/2 if it is an odd integer, meaning n+1 is even
In other words n is odd and S is odd
It is sufficient!!!

Yes (A) is right.
User avatar
ian7777
Joined: 09 Mar 2003
Last visit: 24 Jan 2010
Posts: 227
Own Kudos:
Posts: 227
Kudos: 178
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I go for A, too. If the list had an even number of numbers, then the average would be a .5 (as in, if the list was 4,5,6,7, the average would be 5.5)

If it has an odd number of numbers, then the average is an integer, and if that integer is an odd number, then the sum must have been odd, too.
User avatar
banerjeea_98
Joined: 18 Nov 2004
Last visit: 17 May 2012
Posts: 676
Own Kudos:
Posts: 676
Kudos: 201
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ian7777
I go for A, too. If the list had an even number of numbers, then the average would be a .5 (as in, if the list was 4,5,6,7, the average would be 5.5)

If it has an odd number of numbers, then the average is an integer, and if that integer is an odd number, then the sum must have been odd, too.



That's a good way to figure this one. Avg of even number of +ve consecutive integers is not an integer (rather .5) and for odd number of +ve consecutive integers it is an integer.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
Moderators:
Math Expert
105390 posts
GMAT Tutor
1924 posts