pbajaria wrote:

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

!!Sufficient!!

pbajaria,

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

Now,

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.

B.

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.

- ash

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ash

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I'm crossing the bridge.........