For an imaginary $100 (cheers engage and AGROV8D)

You are given 9 balls, all identical size, color, material and everything else, except for one which weights differently to the others.
Without knowing if the odd ball out is heavier or lighter and using nothing other than a two sided weight scale, in how many weighings could you identify the odd ball?

Shall post answer up at the end of the week and will start tally with the prizes

i would say one

because when i lift up each ball individually, i would be able to tell which one is lighter/heavier when i put them on the scale

where do i collect my imaginary $100?

6 would be the most times you would have to weigh it

5 times
one i would keep one ball in hand, weigh 4 and 4...chances r one side will be heaver and eliminate 4 more

then u would have to weigh each ball after that...

i failed maths lol

4 times. Separate into 4 pairs, with one remaining. Weigh the pairs one at a time. If one is different then you'd notice, else the remaining one would be the different one.

haha im so with drift_boy_007 on this one.......ill say 3 and hope im right and no i cant explain why its 3 cos im guessing.

fuck i dnt even understand the question i failed maths 2

2

hahah you guys failed maths......I didnt even do Maths so where does that put me grrrr!

Haha! yeah! i got about half way through and i started to get a strange sensation in my brain which took me back to my school days... so i stopped!

come on guys uve got 2days left ... blacky is on to it

but you wouldnt know which side has the odd ball, because you dont know if it is heavier or lighter

here is my positive spin on this matter

hahaha that is funny but i think its time for the answer

The answer is ....... 2
Here's how you do it. Divide the balls into 3 groups of 3. Weigh 1st 2 groups of 3 against each other. If they balance, means all the balls weigh the same, so take everything off the scale, take the third group of balls, place 1 ball on the left hand side, one on the right hand side. If they balance, third ball (the one not on the scale) in the group is the heavier one (by process of elimination) or if the scale tips to one side the heavier side is your heavier ball.
If the scale tiped to one side when you weighed the first two groups, you would do the last step for the heavier side and find the heavier ball, hence still taking only 2 weighings to find the heavier ball.
This way coveres all the balls and there are no lucky shots as you can find the heavier ball in two weighings every single time.

So as non of you got it the monies will be added to the next question so the next with be worth $200 imaginary dollars....stay tuned the next question will be up early next week

but you dont kow if the odd ball is heavier or lighter, so you wont know if its the lighter of heavier group off balls, you whould have to know if its heavier or lighter first to do it in 2 goes, unless i have missed something



it would have to be 3 times

only one ball is differnt

so say you weigh the first 2 groups of balls and group 1 is heavier, because you dont know if the odd ball is heavier or lighter you dont know which of the 2 groups of balls the odd one is in, so you will have to weigh the 3rd group of balls to find out.

it duznt matter if the odd one was lighter or hevier if it was differnt then is obvisly gonna be the only one out

Hahahaha!

wow this shit is way 2 hard my head hurts just reading bout it

that doesnt make any sence.

so say you weigh the first 2 groups of 3 balls, and they are the same

you need to then weigh at least 2 of the 3 balls in the third group which would then make your total weighings 3

so your answer is flawed and i still got the imaginary $100 and another bonus $100 imaginary dollars for having to correct your own flawed response to your silly question!