Jul 19 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 20 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 29 May 2008
Posts: 109

If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
19 Aug 2009, 07:35
Question Stats:
46% (02:06) correct 54% (01:56) wrong based on 356 sessions
HideShow timer Statistics
If \(z_1\), \(z_2\), \(z_3\), ..., \(z_n\) is a series of consecutive positive integers, is the sum of all the integers in this series odd? (1) \(\frac{(z_1+z_2+z_3+...+z_n)}{n}\) is an odd integer. (2) n is odd.
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 56260

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
03 Mar 2010, 14:23
swethar wrote: Please solve (with explanation): If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd? 1. [(z1+z2+z3+...zn)/n] is an odd integer. 2. n is odd. Source: KaplanThere is an important property of \(n\) consecutive integers: • If n is odd, the sum of consecutive integers is always divisible by n. Given \(\{9,10,11\}\), we have \(n=3\) consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3. • If n is even, the sum of consecutive integers is never divisible by n. Given \(\{9,10,11,12\}\), we have \(n=4\) consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4. (1) \(\frac{z_1+z_2+z_3+...z_n}{n}=odd\), as the result of division the sum over the number of terms n is an integer, then n must be odd > \(z_1+z_2+z_3+...z_n=odd*n=odd*odd=odd\). Sufficient. (2) \(n\) is odd. Sum can be odd as well as even. Not sufficient. Answer: A. Hope it helps.
_________________




Intern
Joined: 22 Nov 2009
Posts: 23

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
02 Mar 2010, 09:23
If \(z_1\), \(z_2\), \(z_3\), ..., \(z_n\) is a series of consecutive positive integers, is the sum of all the integers in this series odd? (1) \(\frac{(z_1+z_2+z_3+...+z_n)}{n}\) is an odd integer. (2) n is odd.
_________________




Manager
Joined: 14 Aug 2009
Posts: 112

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
19 Aug 2009, 08:06
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. Answer is A.
_________________
Kudos me if my reply helps!



Retired Moderator
Joined: 05 Jul 2006
Posts: 1641

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
Updated on: 06 Sep 2009, 10:52
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
Originally posted by yezz on 06 Sep 2009, 10:49.
Last edited by yezz on 06 Sep 2009, 10:52, edited 1 time in total.



Math Expert
Joined: 02 Aug 2009
Posts: 7764

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
06 Sep 2009, 11:23
ans is A: REASON let S=sum of 'n' consecutive nos with 'a' as first no = n(2a+(n1)d)/2... so S/n= n(2a+(n1)d)/2n=(2a+(n1)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..
_________________



Intern
Joined: 21 Aug 2009
Posts: 27

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



Math Expert
Joined: 02 Aug 2009
Posts: 7764

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
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+(n1)d)/2
_________________



Manager
Joined: 25 Aug 2009
Posts: 153

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
06 Sep 2009, 17:23
Let z1 = k; z2 = k + 1 .....zn = k + n  1
Sum = k + (k+1) + (k +2) +.... + (k+n1) = \(\frac{n(k+k+n1)}{2}\)
=> Sum = \(\frac{n(2k+n1)}{2}\)
1.) Sum/n = odd
=> Sum = n*odd..insufficient..(depends on n.)
2.) n is odd.. Sum = \(\frac{n(2k+n1)}{2}\) \(=> Sum = odd * \frac{(even)}{2}\) 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: 56260

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
06 Sep 2009, 17:49
gmate2010 wrote: Let z1 = k; z2 = k + 1 .....zn = k + n  1
Sum = k + (k+1) + (k +2) +.... + (k+n1) = \(\frac{n(k+k+n1)}{2}\)
=> Sum = \(\frac{n(2k+n1)}{2}\)
1.) Sum/n = odd
=> Sum = n*odd..insufficient..(depends on n.)
2.) n is odd.. Sum = \(\frac{n(2k+n1)}{2}\) \(=> Sum = odd * \frac{(even)}{2}\) 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: 153

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
06 Sep 2009, 17:55
Bunuel wrote: gmate2010 wrote: Let z1 = k; z2 = k + 1 .....zn = k + n  1
Sum = k + (k+1) + (k +2) +.... + (k+n1) = \(\frac{n(k+k+n1)}{2}\)
=> Sum = \(\frac{n(2k+n1)}{2}\)
1.) Sum/n = odd
=> Sum = n*odd..insufficient..(depends on n.)
2.) n is odd.. Sum = \(\frac{n(2k+n1)}{2}\) \(=> Sum = odd * \frac{(even)}{2}\) 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..



Manager
Joined: 26 May 2005
Posts: 183

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
02 Mar 2010, 09:46
swethar wrote: Please solve (with explanation):
If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd? 1. [(z1+z2+z3+...zn)/n] is an odd integer. 2. n is odd.
st 1) [(z1+z2+z3+...zn)/n] is the avg arithmetic mean of the series  for consequetive numbers, if the total number is even, then mean is the avg of the middle two numbers(which is not an interger), or if the total number is odd, then mean is the middle number. As its given mean is an odd interger, the middle number is an odd integer and we will have the same number of positive integers to the right of mean as to the left of mean. and the sum of the remaining integers except mean will be even. ( as for every odd number to the right of mean, there would be an odd number to the left of mean). So the sum of all the numbers in the series is odd. Sufficient st 2) n is odd  sum could be even if the middle number(mean/median) is even [3,4,5] or sum could be odd if the middle number(mean/median) is odd [6,7,8] Not sufficient A



Intern
Joined: 23 Apr 2014
Posts: 11
Location: India

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
27 May 2014, 00:21
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.



Manager
Joined: 22 Jan 2014
Posts: 172
WE: Project Management (Computer Hardware)

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
03 Oct 2014, 06:05
swethar wrote: If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd?
(1) (z1+z2+z3+...zn)/n is an odd integer. (2) n is odd. A. 1) let the numbers be a,a+1,a+2,a+3,...,a+n from FS1 > (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient. 2) n = odd 1+2+3 = 6 (even) 1+2+3+4+5 = 15 (odd) insufficient.
_________________
Illegitimi non carborundum.



Math Expert
Joined: 02 Sep 2009
Posts: 56260

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
03 Oct 2014, 07:33
thefibonacci wrote: swethar wrote: If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd?
(1) (z1+z2+z3+...zn)/n is an odd integer. (2) n is odd. A. 1) let the numbers be a,a+1,a+2,a+3,..., a+n from FS1 > (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient. 2) n = odd 1+2+3 = 6 (even) 1+2+3+4+5 = 15 (odd) insufficient. The last term would be a + n  1, not a + n.
_________________



Manager
Joined: 22 Jan 2014
Posts: 172
WE: Project Management (Computer Hardware)

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
03 Oct 2014, 09:31
Bunuel wrote: thefibonacci wrote: swethar wrote: If z1, z2, z3,..., zn is a series of consecutive positive integers, is the sum of all the integers in this series odd?
(1) (z1+z2+z3+...zn)/n is an odd integer. (2) n is odd. A. 1) let the numbers be a,a+1,a+2,a+3,..., a+n from FS1 > (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient. 2) n = odd 1+2+3 = 6 (even) 1+2+3+4+5 = 15 (odd) insufficient. The last term would be a + n  1, not a + n. Thanks Bunuel. But that would not make the sum odd, still. What am I missing here?
_________________
Illegitimi non carborundum.



Math Expert
Joined: 02 Sep 2009
Posts: 56260

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
03 Oct 2014, 09:55
thefibonacci wrote: Bunuel wrote: thefibonacci wrote: A.
1) let the numbers be a,a+1,a+2,a+3,...,a+n from FS1 > (a+a+1+a+2+...+a+n)/n = odd or (n*a + (n(n+1)/2))/n = odd or (2na + n(n+1))/2n = odd or (2na + n(n+1)) = even*n LHS is nothing bu the sum. n may be even or odd, but RHS would always be even and so would be the LHS. sufficient.
2) n = odd 1+2+3 = 6 (even) 1+2+3+4+5 = 15 (odd) insufficient. The last term would be a + n  1, not a + n. Thanks Bunuel. But that would not make the sum odd, still. What am I missing here? Sorry, but don't know what are you trying to prove there? What's your question? Do you get that n is even? Or ...? If you make last term a + n  1 instead of a + n you'll get that n is odd.
_________________



NonHuman User
Joined: 09 Sep 2013
Posts: 11704

Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is
[#permalink]
Show Tags
05 Jan 2019, 07:02
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 integers, is
[#permalink]
05 Jan 2019, 07:02






