# Brain Teasers are Back!

The 9-ball teaser can be guaranteed to be solved in two weighings.

If we had 27 balls (with one heavy ball) it would be solved in three weighings, whilst 81 would take four weighings. I know that’s heading in the wrong direction but it does show a pattern.

Just trying to give a hint.

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Well, Ravvie has provided quite a hint, and we had better work towards a solution before Mike drowns in whisky !!

In the situation with 9 balls, one of which is heavier …
Put the balls into three groups A, B and C of three balls each group.
Place groups A and B on opposite side of the scales, leaving group C alone.

You should now be able to figure out which group the heavy ball is in.
This is the end of the first weighing !

My outline solution (it works if the number of balls is a power of three) is:

1. After the first weighing, drink a shot of whisky.

2. Repeat until problem solved.

I realise that this is an ambiguous solution which could explain why the thread has gone quiet for several days.

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I’m sticking with my first answer (check the wording of the original teaser).

Best to put the 9-ball teaser to bed. I know Ravvie has the solution.

As I stated a few days ago …

ie if the scales balance, the heavy ball is in the group of three which is not on the scales
If the scales do not balance, the heavy ball is in the group of three on the scales that moved down.

Unload the scales, then using the group of three that contains the heavy ball, place one ball on each side of the scales. If they remain balanced, the heavy ball is the one that isn’t on the scales. If the scales do not balance, the heavy ball is on the scales than moved down.

Solved in two weighings.

As Ravvie has pointed out, 3 balls, 9 balls, 27 balls 81 balls etc require 1 weighing, 2 weighings, 3 weighings, 4 weighings etc, etc to locate the heavy ball.

The situation in the second teaser The odd Ball - upping the anti !! involving 12 balls, one of which is either heavy or light, is a bit more complicated (IMHO).

But can be solved in just three weighings.

The smallest number of weighings possible is 1 though. You can guarantee solving in 2 weighings, but it can sometimes be solved in 1

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I think for 12 balls, the minimum number of weighings is three. … But I’m having a whisky, so could be wrong.

Good whisky Mike, the answer is spot-on at three.
However, your methos needs a bit of work to guarantee a correct solution in three moves

I haven’t had a chance until now to look at the 12 ball teaser. But I am now at an airport at about 2am waiting for an early morning flight so I think my thinking will be as foggy as if I had had a whisky.

The problem with splitting 6 vs 6 is that there is less information available if the first weighing balances. I can’t see a way of solving 6 with just two weighings.

So I think maybe split into 3 lots of 4. The next step would then differ depending on the outcome of the first weighing. I have tried doing it in my head but I got too many outcomes to keep track of.

Yes, I’ve been looking at 3 lots of 4 as well, and even 4 lots of 3, but, like you, I’m struggling to deal with all outcomes.

Steve, Ravvie, you are both on the right track.

I have quoted the teaser using the same words as it was given to me 25/30 years ago. It took me a little while to solve it. However, I assumed that I could recognise which ball was which, even though they all looked the same. Perhaps I should have suggested that the balls are (say) labelled 1 to 12 so that you know which ball is which. You could then start by putting (say) Balls 1, 2, 3, 4 on one side of the scales; 5, 6, 7, 8 on the other side and leave 9, 10, 11, 12 off the scales for the moment.

Apologies if you (and others) consider I should have made this clearer in the initial teaser.

I’ve re-worded this teaser using cubes rather than balls. In fact, it could be a mixture of balls and cubes or even other shapes. The main thing is that one of them is either heavier or lighter than each of the other eleven, which are themselves all of equal of equal weight.

I would group the cubes in 4 sets of 3 : A,B,C and D. Then
compare A and B using the scale.

1. If the weights differ you know that the odd cube is within A or B. Next compare A and C using the scale. The result gives you the odd cube location and whether it is a light cube or a heavy one. Finaly pickup two cubes from the set that contains the odd one, compare their weights. The result gives you the answer of the problem since you know whether the odd cube is a heavy one or a light one. It takes 3 moves.
2. the case where A and B have the same weights is solved using the same method. But it may take 4 moves.

That’s a really nice attempt Alain.

I do like your description in paragraph No.1 Very clear. However …
… as you rightly anticipate in paragraph No.2, it will take four or more moves if A and B turn out to have the same weights.

The problem can guarantee to be solved in only 3 moves, but does require a somewhat different approach.

But as I said above, really nice attempt.

Great teaser which proved too difficult for me. After a while gave up and resorted to the internet and amazingly found 3 solutions – all of which I’d never have thought of myself!! Good luck, folks.

Hint: Numbering the cubes 1 through to 12 and splitting into 3 groups of 4 is one way forward.

Thanks Cluffy, I encountered the teaser some 25 or 30 years ago, before the internet was up and running so to speak.

Your hint is a good start for many, and is the way I started my sloution all those years ago. I was actually on a delayed train and had plenty of time. But it still took a flash of inspiration to visualise a solution.

And a real gent you are too, to declare your use of the internet. Kudos !

I propose the following:

Put balls 1 to 4 on one side of the scale, balls 5 to 8 on the other.

1. If they have the same weight, you deduce that the odd ball is among balls 9 to 12.
a) put balls 9 and 10 on one side of the scale, and 11 and 12 on the other
b) put balls 9 and 11 on one side of the scale, 10 and 12 on the other
Combining these two measurements gives you the result

2. If they do not have the same weight, put 1,2, 5 and 6 on one side of the scale, 3,4, 7 and 8 in the other (namely permute 3 and 4 with 5 and 6).
a) If the scale gives the same result you deduce that the odd ball is not among balls 3,4,5 and 6. You finally put 1 and 7 on one side of the scale, 2 and 8 on the other. Combining these two measurements gives you the result
b) If the scale gives a different result you deduce that the odd ball is not among balls 1,2 7 and 8. (…)

It seems like we are making progress but I don’t think we are quite there yet.

In 1, suppose 9 is heavy. Then scale A will drop for both (a) and (b). Alternatively, suppose 12 is light. Then scale A will drop for both (a) and (b), as before. Hence after 3 weighings we can’t be sure of the result.

I don’t have an answer yet though.

I might give this a go……

A bit of encouragement …

My solution starts with 3 groups each with 4 balls eg (1, 2, 3, 4) (5, 6, 7, 8) and (9, 10, 11, 12)

Put the first two groups, one each side of the balance. (9, 10, 11, 12) are not on the scales.

If the scales balance, the odd ball is one of 9, 10, 11 or 12 AND
Balls 1 to 8 are all standard weight balls.

Try working through the rest of this part of the process to a solution. (this is the easier part)

Then return to … If the scales DO NOT balance, the odd ball is one of either (1, 2, 3, 4) or (5, 6, 7, 8) …