Okay heres a logic puzzle i was given and i've spent weeks trying to figure out...?

Question

Okay heres the problem and it's been blowing my mind.

There are 12 balls. All of them are exactly the same shape, size, texture (Exactly the same) apart from one of them. You can't tell which one just by holding them you have to weigh them on scales.

The ball could be lighter than the other 11 or heavier than the other 11.
The problem is you have to use the weighing scales but can only weigh the balls 3 times.
From weighing the balls you have to determine
1. which ball is different?
2. Is the ball heavier or lighter?

apparently there is no trick to the question like how many eggs can a rooster produce?

Answer 1

I haven't spent weeks on it, but I can't get it either I'm afraid. If I had 4 uses of the scale, or if I knew that the oddball was heavier or lighter I could do it. I'd be intrigued to know the answer.

Sorry Naveen, there are 12 balls, not 9.

OK, I couldn't help myself. I cheated and looked up the answer. You were given a very difficult puzzle to solve. You can see that after reading the answer on this link http://blogs.sun.com/dcb/resource/SOLUTION_12_Balls.html

It seems confusing at first, but once you read through it a few times you'll realize the author is correct. If you want to wait till someone figures it out here go ahead, otherwise you can use my cheat ;-)

Answer 2

put all balls on the scale together, and remove them 1 at a time, you will soon see which the odd one out is by process of elimination if you note what weight is reduced by each time you remove a ball, when the odd one is picked the weight will either be lighter or heavier, and that way you are only weighing the balls once. as a guess i would say lighter.

Answer 3

Well, if it's slighty heavier OR lighter than that will throw off my whole answer. But here goes. First, weigh 6 marbles on each side. If you're looking for the heavier marble, then you know it's on the side that weighs the most. So split those 6 to weigh 3 on each side, and again determine the heaviest pile. With the remaining 3, weigh 1 against 1, and set the 3rd one aside. If the two you chose to weigh weigh the same, then heavier marble is the one you set aside. If one is heavier, then obviously that one would be the marble. And this whole things works same if you're looking for the lighter marble. But I have no idea how to do it looking for an unknown either heavier or lighter. If you have the answer, let me know after this question's over how to do it. =)

Answer 4

Initially weigh six balls three balls on each side

---------1) If they are equal then the fake ball is among three that is not weighed .now weigh any two balls from the remaining three if they are equal the fake ball is one that is not weighed ,if not equal then replace one ball if both are equal the fake ball is in ur hand , if it is not equal then the fake ball is the one that is not changed (took only three chances)

--------2) If they are not equal replace the three balls with the remaining three by this we can find the set in which the fake ball is present and also whether it is heavier or lighter . Then weigh two balls from the three if equal the fake ball is in ur hand else based on whether it is heavy or light v can identify the ball(now also only 3 chances)

Answer 5

Weigh all of the balls on the scale, the lighter half you weigh again (if it is equal that means that it was the heaver one... oops) repeat process until you get the lightest ball. Have fun!

Answer 6

divide the balls 3 piles of 4 balls and weigh em.
One pile will be different in weight, the one that contains the odd ball.
Divide that pile into 2 and weigh. from the weight of the matching piles you should be able to deduct which pair of balls contains oddball.
Now you've 2 balls and i've just lost my train of thought.
this is going to be winding me up too now!!!

Answer 7

Go for the Ball number 7 it'll be heavyer than the others.