Evaluating choice 1
1) if choice 1 is true then numerator and denominator are odd.
odd/odd = odd (15/5,45/9, etc)
only exception is n/n = 1 which is not possible in this case since numeration is sum of n numbers
I just wanted to know how you concluded that both Numberator and Denominator is odd from your explanation above.
6/2 = 3
here Nr and Dr are both even and the quotient is odd.
Now , here's how i concluded that n is odd.
Given that Expression E = Sum(Z1...Zn)/n = odd
Sum(Z1...Zn) = n[Z1+Zn]/2
E = [Z1+Zn]/2 = odd
therefore Z1+Zn = 2*odd = even
from this we know that both Z1 and Zn are either odd or even.
If n is even Z1 and Zn will never be both odd or both even.
Only if n is odd, Z1 and Zn are both odd or both even and then only can their sum be even.
So the series can only be one of the following types
1 2 3
2 3 4 etc
i.e. n has to be odd. A is sufficient.
You can pick numbers and solve it faster, but if you know the concept
you have an upper hand, just in case picking numbers doesn't workout sometime.
In this case I picked numbers.
given n is odd.
let the series be 123. sum = 6
let the series be 234. sum = 9
So B is insufficient.
What I generally do is like this. If I have to prove something is always true then I try to use the algebraic and arithmetic concepts.
When I dont have to prove that something is always true, I pick numbers.
I'm crossing the bridge.........