## The weighing twelve (12) balls riddle – which one is a different weight?

This is a riddle that a friend (Carl “Chunky” Benson) put to me a few years back. A few A4 pages of scribblings and I had accounted for an “answer” for *almost* all possible outcomes (almost), however, I never arrived at a complete solution.

There’s every chance that I’m just hopeless at Maths, so prove me wrong people

Premise

- You have
**twelve (12)** balls and a set of balance scales.
**One** **(1)** of the balls is a *different* weight to the other **eleven (11)** balls.
- You are allowed to use the balance scales
**three (3) **times.
- You need to determine which ball is the “odd one out” and whether it is heavier or lighter than the other balls.

Now these are *balance* scales I’m talking about. While a solution involving electronic scales would be effective, I don’t think it’s what the riddle’s architect had in mind

Balance Scales

I’ve seen a few similar riddles such as this posted elsewhere on the Internet, however, several state that the ball is either *heavier* or *lighter*; a far easier problem to solve. Maybe it should be called something different than “weighing twelve…” as your solution mightn’t necessarily involve weighing all the balls.

Good luck!

Feel free to add any comments. If you require further clarification on any part of this riddle, please leave a comment, or you can email me at: ryan@kirgs.com

*Ryan Kirgan lives in and works out of Sydney, Australia.*

Ryan Kirgan

P.S – Any hate mail you want to write (after obsessing over this problem) can be sent to carl@mytrade.com.au

## { 12 } Comments

I’ve been told by Carl that it was coins rather than balls, however the premise remains the same.

I’m sure if you Google Search “12 coins” you’ll get your result (and may save some heart/headache in the process!)

The key to this is essentially base-three counting. Each ball gets labeled with a unique 3-digit base three number. The digits tell us where to put the ball for each weighing (e.g. 0 means not weighed, 1 means put it on the left, 2 means put it on the right).

The idea is then that we do the weighings, and the results tell us the label of the odd ball. So if the weighings were: left pan heavy, right pan heavy, balance, then the odd ball is 120

However, since we don’t know whether the ball is heavy or light, each ball has to get two labels which are opposites of each other (so 120 would also have 210).

There are twenty-seven possible labels, but we leave out 000, 111, and 222. This leaves 24, which when paired up gives us exactly twelve sets. These sets have the property that the balls are split up 4-4-4 for each weighing.

What I’ve written isn’t a proof, but it should get you started. At the very least, it is an easy way to remember the solution to this problem.

The first (external) post to this blog entry (in fact, the first external post to this blog) is absolutely brilliant!

Your very articulate response is much appreciated and I look forward to trying the method out!

Thanks.

With a balance scale, this problem is fairly easy. However, I was more intrigued by the idea of an electronic scale (to figure weight). Is it possible for this riddle to be done with a weighing devices and not a balance device, in only 3 tries? Just trying to figure it out using that method.

@Joseph Demke

This is a really good point – I just assumed it would be easier, but come to think of it the balance scales really do give you more information.

Back to the drawing board

There are 12 poker chips and one of them is different in weight than the others. Using only a calibrated balance scale, how can you determine the odd chip in only 3 weighings?

- Step 1: Balance 2 piles of 4 chips

- Case 1: Both Piles Level

o Step 2: Take 3 of the 4 unknown chips and weigh against 3 of the known normals

Case 1: Both Piles Level

• Step 3: Take the remaining chip and weigh against a known normal to get answer

Case 2: Piles not level

• Take the 3 potentially light or heavy chips (you know which one they are now) and balance two of them to see which is the lighter or heaver one or if it’s the one in your hand.

- Case 2 (Worst Case): Piles not Level

o Identify the 4 potentially heavy and 4 potentially light chips from step 1

o Step 2

Side 1 of scale: 3 potentially light chips and 1 of the potentially heavy chips

Side 2 of scale: 3 of the 4 known normal chips from step 1 and the 1 leftover potentially light chip

Case 1: Both Piles Level

• Step 3: Take the remaining 3 potentially heavy chips and balance 2 of them, seeing which is the heaver one or if it’s the 3rd one in your hand

Case 2: Side 1 is higher

• Step 3: As none of the chips on side 2 can be heavy, it must be one of the 3 potentially light chips on side 1, so do like step 3 above for them to determine the light chip

Case 3: Side 2 is higher

• Step 3: Since there is only one heavy chip on side 1 and one light chip on side 2, we know that one of them is the culprit, so take one of them and weigh it against a known normal chip and get your answer!

I came to this conclusion by thinking about how I wanted to have 3 chips at the end that were either all potentially heavy or potentially light. This made me want to eliminate 5 of the 8 unknowns in step 2. The key was having only one potentially heavy chip on side 1 and only one potentially light chip on side 2 so that if side 2 becomes higher, it must be one of those two chips. I kinda lucked out picking groups of 4 for step one, but I spent considerable time examining other groupings for step1 and 3 groups of 4 seemed to be getting the closest.

-Chaz Morantz

1) Weigh 4 and 4 assume they are not even.

2) Mark all as either 1-4(+) 5-8(-) or 9-12(E) *E for Even

3) Weigh 5+1-9E against 6+2-7+

If Even –> Weigh the two remaining 3-4- (if uneven take the lighter one)

If Left is heavier –> weigh 5+E

If Right is heavier –> weight 6+7+ (if uneven take the heavier one)

Put 6 balls or coins on each side. Then take the heavier or lighter side ad divide into tho piles of 3. Weight those piles and take the heavier or lighter side. Now weigh any 2 of the balls or coins on each side. If the two balances, the extra one is your answer. Otherwise you will get your answer on the scale.

These other solutions is way too complicated for my brain to work out!

I missed one part of the riddle. Let me think about it for a while! Ouch!

all the answers in the comments are wrong… this riddle is fairly basic and simple.

3 balls on each side of the scale. if they are even..throw them away.

now you have 6 remaining balls.

same move again, the heavier side of the scale has the heavy ball and you are left with 1 more try on the scale.

1 ball on each side, if they even out, the 1 left out is the heavy 1

Hisham, there are a couple of issues with your response, but an important aspect is that you don’t know if the unique ball is heavier or lighter. There are a couple of methods to solving this, but it is a bit more complex.

Sumudu’s approach below is a good one and the best for mine.

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