If z1, z2, z3,..., zn is a series of consecutive positive : GMAT Data Sufficiency (DS)
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# If z1, z2, z3,..., zn is a series of consecutive positive

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

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02 Mar 2010, 08:23
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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

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Re: DS - Sum of integers [#permalink]

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02 Mar 2010, 08:46
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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.

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

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Re: DS - Sum of integers [#permalink]

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03 Mar 2010, 13:23
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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.

[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.

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

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03 Oct 2014, 00:53
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Re: If z1, z2, z3,..., zn is a series of consecutive positive [#permalink]

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03 Oct 2014, 05:05
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 [#permalink]

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03 Oct 2014, 06:33
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 [#permalink]

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03 Oct 2014, 08:31
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 [#permalink]

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03 Oct 2014, 08:55
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 [#permalink]

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09 Jul 2016, 16:15
Hello from the GMAT Club BumpBot!

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

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09 Jul 2016, 21:53
A

take cases 1,2,3,4,5 or 2,3,4,5,6

these tow cases will solve this question.
Re: If z1, z2, z3,..., zn is a series of consecutive positive   [#permalink] 09 Jul 2016, 21:53
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