Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 27 May 2017, 06:21

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If z1, z2, z3, ..., zn is a series of consecutive positive

Author Message
TAGS:

### Hide Tags

Manager
Joined: 29 May 2008
Posts: 113
Followers: 1

Kudos [?]: 102 [0], given: 0

If z1, z2, z3, ..., zn is a series of consecutive positive [#permalink]

### Show Tags

19 Aug 2009, 07:35
7
This post was
BOOKMARKED
00:00

Difficulty:

85% (hard)

Question Stats:

40% (02:39) correct 60% (01:48) wrong based on 136 sessions

### HideShow timer Statistics

If z1, z2, z3, ..., zn is a series of consecutive positive integers, is the sum of all integers in this series odd?

(1) (z1+z2+z3+...+zn)/ n is an odd integer
(2) n is odd
[Reveal] Spoiler: OA
Manager
Joined: 14 Aug 2009
Posts: 123
Followers: 2

Kudos [?]: 105 [1] , given: 13

### Show Tags

19 Aug 2009, 08:06
1
KUDOS
TheRob wrote:
Hi , I have always had trouble with this thype of questions, would you please explain me how to solve it and how to get better at it?

If z1,z2,z3,...,zn is a series of consecutive positive integers, is the sum of all integers in this series odd?

1) (z1+z2+z3+...+zn)/ n is an odd integer

2) n is odd

for (2), suppose n=3,
if z={1,2,3}, sum(z)=6
if z={2,3,4}, sum(z)=9
therefore 2) is nsf.

for (1),
(z1+z2+z3+...+zn)/ n =m, and m is an odd integer, therefore n is odd
consequently, sum(z)=n*m is an odd figure.

_________________

Kudos me if my reply helps!

Manager
Joined: 29 May 2008
Posts: 113
Followers: 1

Kudos [?]: 102 [0], given: 0

### Show Tags

19 Aug 2009, 12:03
How do you find out that n is odd?
SVP
Joined: 05 Jul 2006
Posts: 1747
Followers: 6

Kudos [?]: 358 [0], given: 49

### Show Tags

06 Sep 2009, 10:49
TheRob wrote:
Hi , I have always had trouble with this thype of questions, would you please explain me how to solve it and how to get better at it?

If z1,z2,z3,...,zn is a series of consecutive positive integers, is the sum of all integers in this series odd?

1) (z1+z2+z3+...+zn)/ n is an odd integer

2) n is odd

for the sum to be odd, n = odd , odd numbers included are odd in number

from 1

sum is even or odd., n is odd OR EVEN..........insuff

from 2

insuff

BOTH

n is odd still sum is surely odd ........C

Last edited by yezz on 06 Sep 2009, 10:52, edited 1 time in total.
Verbal Forum Moderator
Joined: 02 Aug 2009
Posts: 4514
Followers: 394

Kudos [?]: 4201 [0], given: 109

### Show Tags

06 Sep 2009, 11:23
Expert's post
1
This post was
BOOKMARKED
ans is A: REASON
let S=sum of 'n' consecutive nos with 'a' as first no = n(2a+(n-1)d)/2...
so S/n= n(2a+(n-1)d)/2n=(2a+(n-1)d)/2, which is equal to the avg of n nos....

now in consecutive nos avg is the center no if n is odd or avg of center two nos,which would be in decimals((odd + even)/2) if n is even..
by statement I...avg of n consecutive nos is an odd no... therefore n is odd .. so sum is odd no *odd no= odd no..
hence sufficient..
II is not sufficient..
i hope it was of some help to those asking how n is odd..
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Intern
Joined: 21 Aug 2009
Posts: 41
Followers: 1

Kudos [?]: 34 [0], given: 5

### Show Tags

06 Sep 2009, 12:10
Is Sn=n(n+1)/2 odd ?; Sn= sum of consecutive positive no.s

stat1: Sn/n is odd => n(n+1)/2n is odd or (n+1)/2 is odd
or (n+1) is even and => n is odd or Sn = n(odd) * (n+1)/2 (odd) = odd suff.

stat2: n is odd => n+1 is even and n(n+1) is even or Sn = n(n+1)/2 is even... suff.

IMO D
Verbal Forum Moderator
Joined: 02 Aug 2009
Posts: 4514
Followers: 394

Kudos [?]: 4201 [0], given: 109

### Show Tags

06 Sep 2009, 12:23
Is Sn=n(n+1)/2 odd ?; Sn= sum of consecutive positive no.s

stat1: Sn/n is odd => n(n+1)/2n is odd or (n+1)/2 is odd
or (n+1) is even and => n is odd or Sn = n(odd) * (n+1)/2 (odd) = odd suff.

stat2: n is odd => n+1 is even and n(n+1) is even or Sn = n(n+1)/2 is even... suff.

ANS:-
i think u r going wrong on stat2...
eg if 3 nos are 1,2,3.. n is 3 ie odd however its sum is 6 which is even..
or 3 nos are 2,3,4.. n is 3 ie odd however its sum is 9 which is odd..
so not sufficient
Sn=n(n+1)/2 is the sum of first consecutie positive nos .....here it is not given that they are first consecutie positive nos but only that they are consecutie positive nos, where Sn = n(2a+(n-1)d)/2
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Manager
Joined: 25 Aug 2009
Posts: 175
Followers: 1

Kudos [?]: 91 [0], given: 12

### Show Tags

06 Sep 2009, 17:23
Let z1 = k; z2 = k + 1 .....zn = k + n - 1

Sum = k + (k+1) + (k +2) +.... + (k+n-1) = $$\frac{n(k+k+n-1)}{2}$$

=> Sum = $$\frac{n(2k+n-1)}{2}$$

1.) Sum/n = odd

=> Sum = n*odd..insufficient..(depends on n.)

2.) n is odd..
Sum = $$\frac{n(2k+n-1)}{2}$$
$$=> Sum = odd * \frac{(even)}{2}$$
Now, Even/2 can be odd or even..we can not be sure..insufficient..

combining both..

n is odd..

=> Sum is odd ..hence, C
Math Expert
Joined: 02 Sep 2009
Posts: 38911
Followers: 7741

Kudos [?]: 106305 [0], given: 11620

### Show Tags

06 Sep 2009, 17:49
Expert's post
1
This post was
BOOKMARKED
gmate2010 wrote:
Let z1 = k; z2 = k + 1 .....zn = k + n - 1

Sum = k + (k+1) + (k +2) +.... + (k+n-1) = $$\frac{n(k+k+n-1)}{2}$$

=> Sum = $$\frac{n(2k+n-1)}{2}$$

1.) Sum/n = odd

=> Sum = n*odd..insufficient..(depends on n.)

2.) n is odd..
Sum = $$\frac{n(2k+n-1)}{2}$$
$$=> Sum = odd * \frac{(even)}{2}$$
Now, Even/2 can be odd or even..we can not be sure..insufficient..

combining both..

n is odd..

=> Sum is odd ..hence, C

(1) S=n*odd, S can be odd or even - generally right. But here, we have consecutive positive integers and here if S is even average is always decimal, if S is odd average can be even or odd, so if average=S/n is not decimals already means that S is odd.

So A
_________________
Manager
Joined: 25 Aug 2009
Posts: 175
Followers: 1

Kudos [?]: 91 [1] , given: 12

### Show Tags

06 Sep 2009, 17:55
1
KUDOS
Bunuel wrote:
gmate2010 wrote:
Let z1 = k; z2 = k + 1 .....zn = k + n - 1

Sum = k + (k+1) + (k +2) +.... + (k+n-1) = $$\frac{n(k+k+n-1)}{2}$$

=> Sum = $$\frac{n(2k+n-1)}{2}$$

1.) Sum/n = odd

=> Sum = n*odd..insufficient..(depends on n.)

2.) n is odd..
Sum = $$\frac{n(2k+n-1)}{2}$$
$$=> Sum = odd * \frac{(even)}{2}$$
Now, Even/2 can be odd or even..we can not be sure..insufficient..

combining both..

n is odd..

=> Sum is odd ..hence, C

(1) S=n*odd, S can be odd or even - generally right. But here, we have consecutive positive integers and here if S is even average is always decimal, if S is odd average can be even or odd, so if average=S/n is not decimals already means that S is odd.

So A

hmmm..i forgot to apply the property of sum of consecutive integers...Thanks for correcting me..
Intern
Joined: 21 Aug 2009
Posts: 41
Followers: 1

Kudos [?]: 34 [0], given: 5

### Show Tags

07 Sep 2009, 08:29
Agree, I made a mistake; it should be A
Intern
Joined: 23 Apr 2014
Posts: 11
Location: India
Followers: 0

Kudos [?]: 10 [0], given: 46

Re: If z1, z2, z3, ..., zn is a series of consecutive positive [#permalink]

### Show Tags

27 May 2014, 00:21
2
This post was
BOOKMARKED
TheRob wrote:
If z1, z2, z3, ..., zn is a series of consecutive positive integers, is the sum of all integers in this series odd?

(1) (z1+z2+z3+...+zn)/ n is an odd integer
(2) n is odd

As per the information

1. Average of number from Z1 to Zn is an odd integer.

In other words, (First Term+Last Term)/2= odd integer
FT+LT= Even integer

It is only possible if FT and LT are both odd or both even at the same time.

Now between x consecutive positive odd integers the number of terms is odd
between x consecutive positive even integers the number of terms is odd

so the sum of the terms= [(FT+LT)/2] X no. of terms from FT to LT inclusive
= odd integer X odd integer
= odd integer
1. Sufficient

2. Clearly not sufficient.

Ans. A.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15480
Followers: 651

Kudos [?]: 209 [0], given: 0

Re: If z1, z2, z3, ..., zn is a series of consecutive positive [#permalink]

### Show Tags

13 Jul 2016, 05:46
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If z1, z2, z3, ..., zn is a series of consecutive positive   [#permalink] 13 Jul 2016, 05:46
Similar topics Replies Last post
Similar
Topics:
5 If p, q, r are a series of three consecutive positive integers, is the 6 26 Nov 2014, 07:19
5 If x, y, and z are 3 positive consecutive integers such that x < y < z 5 20 Mar 2017, 22:21
26 Given a series of n consecutive positive integers, where n > 8 22 May 2016, 04:35
15 If z1, z2, z3,..., zn is a series of consecutive positive integers, is 9 09 Jul 2016, 22:53
4 For positive integer Z, is the expression (Z + 2)(Z^2 + 4Z + 5 29 Jul 2014, 01:06
Display posts from previous: Sort by