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

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

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New post 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.

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

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New post 06 Sep 2009, 11:23
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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..
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Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is [#permalink]

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New post 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

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

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

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New post 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

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

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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
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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..

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

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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.
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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

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

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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.

[Reveal] Spoiler:
OA A


Source: Kaplan


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

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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.

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

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

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

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

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New post 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.
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Re: If z1, z2, z3,..., zn is a series of consecutive positive integers, is   [#permalink] 15 Sep 2017, 04:02
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If z1, z2, z3,..., zn is a series of consecutive positive integers, is

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