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# if z1,z2,z3, ... zn is a series of consequtive positive

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VP
Joined: 25 Nov 2004
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if z1,z2,z3, ... zn is a series of consequtive positive [#permalink]

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27 Jan 2005, 00:18
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ok guys, we are having short of problems. let us post some problems.......

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

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

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SVP
Joined: 03 Jan 2005
Posts: 2228

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Re: DS: odd or even??????????? [#permalink]

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27 Jan 2005, 00:48
MA wrote:
ok guys, we are having short of problems. let us post some problems.......

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

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

If n is even, then there's always even numbers of odd numbers and thus the sum would be even
If n is odd, however, depends on if the starting number is odd or even.
(i)S/n is odd
S=S/n*n if n is even, S is even; if n is odd, S is odd.
Insufficient
(II) n is odd
insufficient as it depends on the starting number

Combine together, we can determine that S is odd. Sufficient

(C)

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Manager
Joined: 11 Jan 2005
Posts: 101

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27 Jan 2005, 01:20
per definition

odd/odd = odd
even/odd = even

odd + even = odd
odd + odd = even
finally odd + even + odd = even
as such

i. no sufficient
ii. no sufficient, we know only that n is odd

both : sufficient
as we know that the division by n (as odd) is odd
so the sum is odd

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Manager
Joined: 06 Sep 2004
Posts: 69

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27 Jan 2005, 06:19
(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.

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SVP
Joined: 03 Jan 2005
Posts: 2228

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27 Jan 2005, 09:08
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.

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CIO
Joined: 09 Mar 2003
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27 Jan 2005, 10:00
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.

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VP
Joined: 18 Nov 2004
Posts: 1430

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27 Jan 2005, 10:13
ian7777 wrote:
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.

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27 Jan 2005, 10:13
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