I give you 12 ball-bearings. They look and feel identical. I tell you that each ball-bearing weighs exactly the same, except for ONE of them, which is a slightly different weight (but I don't tell you whether it's heavier or lighter - it could be either heavier or lighter than the others, and you don't know which).

You have a set of scales (the old-fashioned type that has two bowls on either side of a central pivot).

In just three weighings of the balls (with the scales), you must identify the "odd" ball and tell me whether it is lighter or heavier than the others.

It's not easy - but it can be done!

(If you already know the answer, hold fire while the others think about it).

Does this work?

Say the balls are numbered 1 to 12

First weighing
1,2,3,4 vs 5,6,7,8

If they are evenly balanced, then you know it is 9,10,11 or 12

...So you then weigh 9,10 vs 1,2

......If they are evenly balanced, then you know it is 11 or 12
.........So you weigh 11 vs 1
.........You then know whether it is 11 or 12

......If they are not evenly balanced then you know it is 9 or 10 and you weigh 9 vs 1 and proceed as above

If the first balance was not even, then you know it is 1-8

Second Weighing

1,2,5 vs 3,6,7

...If they are evenly balance, you know it is 4 or 8. You can find out which by weighing 4 vs 1

...If they are not evenly balanced then if you weigh 1,5 vs 2,7

......If they are evenly balanced then you know it was 6 or 3. You know which because if it was 3 it would have tipped one way first time and the other way second time. You therefore know whether it is 6

.....If they are not evenly balanced then - if it tipped LLL - it is 1, if it was LLR - it was 2, if it was LRL - it was 7 and if it was LRR it was 5.

THe only problem is that I do not think you can always tell whether the ball is heavy or light

Am I along the right lines???

Yes - you are on the right lines, but it IS possible to always identify the odd ball AND to say whether it is heavier or lighter. It takes a while to solve this. It's not straightforward, but it is satisfying when you finally get to the answer. Keep going!

Oooh I remember this one..... very nice solution!!!!!

OK - here is a solution that works

First weigh
1,2,3,4 vs 5,6,7,8

Second weigh
1,2,5,9 vs 3,6,10,11

Third weigh
1,5,8,11 vs 2,7,10,12

The following 24 outcomes are possible (U - means right side up; D - means right side down; - - means flat; H - means heavy; L - means light)

1H - UUU
1L - DDD
2H - UUD
2L - DDU
3H - UD-
3L - DU-
4H - U--
4L - D--
5H - DUD
5L - UDU
6H - DD-
6L - UU-
7H - D-D
7L - U-U
8H - D-U
8L - U-D
9H - -U-
9L - -D-
10H - -DD
10L - -UU
11H - -DU
11l - -UD
12H - --D
12L - --U

Pxs - your method looks ingenious, although I must admit I have not yet checked it through. If it's right, then well done!

Will post my solution shortly (which is rather different).

Light, same, heavy ??
got some nice base 3 arithmetic there.
three weighings, three outcomes per weighing = 27 distinct states of the system. Given that ONE of the balls is either Heavier or Lighter... it looks about right that some of the outcomes are invalidated so the state table is reduced to the right size. Impressive.

Very nice pxs.

I think we have a genuis disguised as a magician in here (pxs) very good job.

This is the solution I came up with.

First:
1 2 3 4 5 6 and 7 8 9 10 11 12

Lets say 7 8 9 10 11 12 is heavier.

Second:
7 8 9 and 10 11 12

Lets say 7 8 9 is heavier.

Third:
7 and 8

Now you know which is heavier. If they are the same, then 9 is the heavier one.

But what if it is bearing number 3 that is lighter? 7 8 9 10 11 & 12 would still be heavier.

I remember getting this in a maths exam at school, it was kind of like a test which our teacher gave us to see whether we could do it. Great puzzle, thanks for reminding me.
Owen

This is indeed a great puzzle. Here are a couple of interesting links...

The OddBall Problem

Odd Coin Problems