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z1,z2,z3....zn is a series of consecutive +ve integers; is

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z1,z2,z3....zn is a series of consecutive +ve integers; is [#permalink] New post 21 Oct 2005, 00:37
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A
B
C
D
E

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z1,z2,z3....zn is a series of consecutive +ve 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

Kaplan - can someone explain in detail also pl?
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 [#permalink] New post 21 Oct 2005, 07:30
1. (z1+z2+z3+..+zn)/ n is an odd integer

Insuff: Sum could be 9 and n could be 3 or Sum could be 12 and n could be 4.

2. n is odd

Insuff: Sum could be 9 or 12.

Combining,

Sum = n * average
= odd (stmt 2) * odd (stmt 1)
= odd

Hence answer is C
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 [#permalink] New post 21 Oct 2005, 09:47
A

A) Sum/n can be odd only if sum and n are both odd or both even.
If n is even, the average of n numbers is not an integer. So both sum and n are odd.

B) n=odd.
{2,3,4} = 9/3 = odd
{3,4,5} = 12/3 = even
Insuff.
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 [#permalink] New post 21 Oct 2005, 10:04
yup... A

gsr's method is correct.
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 [#permalink] New post 21 Oct 2005, 10:14
gsr wrote:
A

A) Sum/n can be odd only if sum and n are both odd or both even.
If n is even, the average of n numbers is not an integer. So both sum and n are odd.
.

I just want to add something: when even/even
the sum= n/2 *( z1+zn)
average= [(n/2)*(z1+zn)]/n= (z1+zn)/2
when n is even (z1+zn) is always odd thus the average is not an integer as gsr pointed out.
  [#permalink] 21 Oct 2005, 10:14
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