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**cnaumann****Member**- Registered: 2005-05-10
- Posts: 2

A friend of mine told me this problem, and I worked out a solution. Then I looked up the solution on this website. I like my solution much better!

The Problem:

You have 12 balls identical in size and appearance but 1 is an odd weight (could be either light or heavy).

You have a set scales (balance) which will give 3 possible readings: Left = Right, Left > Right or Left < Right (ie Left and Right have equal weight, Left is Heavier, or Left is Lighter).

You have only 3 chances to weigh the balls in any combination using the scales. Determine which ball is the odd one and if it's heavier or lighter than the rest. How do you do it?

My Solution:

Label the balls 1-12

First Weighing:

Left: 1 2 3 4

Right: 5 6 7 8

Off: 9 10 11 12

Record the heavier side (L, R, or B)

Second Weighing:

Left: 1 2 5 9

Right: 3 4 10 11

Off: 6 7 8 12

Record the heavier side (L, R or B)

Third Weighing:

Left: 3 7 9 10

Right: 1 4 6 12

Off: 2 5 8 11

Record the heavier side (L, R, B)

There are 27 (3^3) possible combination of scale readings. A complete sorted list of the scale reading appears below. Note that only 24 of the 27 readings should be possible given the original problem statement. The algorithm was designed so that if all three scale readings are the same, an error is flagged indicating that the scale is stuck.

BBB Error! There is not a single light or heavy ball (or scale is stuck).

BBL Ball #12 is light

BBR Ball #12 is heavy

BLB Ball #11 is light

BLL Ball #9 is heavy

BLR Ball #10 is light

BRB Ball #11 is heavy

BRL Ball #10 is heavy

BRR Ball #9 is light

LBB Ball #8 is light

LBL Ball #6 is light

LBR Ball #7 is light

LLL Error! Scale is stuck!

LLB Ball #2 is heavy

LLR Ball #1 is heavy

LRB Ball #5 is light

LRL Ball #3 is heavy

LRR Ball #4 is heavy

RBB Ball #8 is heavy

RBL Ball #7 is heavy

RBR Ball #6 is heavy

RLB Ball #5 is heavy

RLL Ball #4 is light

RLR Ball #3 is light

RRB Ball #2 is light

RRL Ball #1 is light

RRR Error! Scale is stuck!

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,654

What a neat approach ... I haven't checked your answer, but the approach looks sound !

I will put it as part of official solution when I get a chance. Should I credit "cnaumann" ?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**cnaumann****Member**- Registered: 2005-05-10
- Posts: 2

Thanks! I am glad you liked my solution. You can credit Charles Naumann.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,654

I have added your solution here.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Mr T****Member**- Registered: 2005-03-30
- Posts: 1,012

wow ur famous

I come back stronger than a powered-up Pac-Man

I bought a large popcorn @ the cinema the other day, it was pretty big...some might even say it was "large

Fatboy Slim is a Legend

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**tt****Member**- Registered: 2005-07-03
- Posts: 32

Hello. I'm a new guy in here. Mr. Cnaumann 's solution is so impressive and systematic. Bravo !

I figured out a solution several months ago, although less nimble but...I tried my best :

Weigh (1st time) 4 balls against other 4 (choose randomly).

IF THE SCALE NOT BALANCED: the last 4 balls (not on the scale) are normal ones.

Name the 4 balls on heavier side A,B,C,D. If the odd ball is in group ABCD, it must be heavier than normal one.

Name the 4 balls on lighter side E,F,G,H. If the odd ball is in group EFGH, it must be lighter than normal one.

Weigh (2nd time) : (N is a normal ball)

A B N ~ C D E

1a/ If ABN = CDE : the odd ball is in group F,G,H.

Weigh F against G (3rd time):

If not balanced, the lighter one is the odd ball.

If balanced, H is the odd ball (and is lighter).

1b/ If ABN ≠ CDE :

1b1/ If the scale tipped towards A,B,N side (heavier): the odd ball is in group A,B,E.

Weigh A against B (3rd time) :

If not balanced, the heavier one is the odd ball.

If balanced, E is the odd ball and lighter.

1b2/ If the scale tipped towards C,D,E side : the odd ball is in group C,D.

Weigh C against D. The heavier one is the odd ball.

IF THE SCALE BALANCED (at the 1st weighing) : these 8 balls are normal ones.

Name the last 4 balls A,B,C,D.

Weigh (2nd time) :

A N ~ C D

1a/ If not balanced (AN≠ CD) : the odd ball is in group A,C,D.

1a1/ If the scale tipped towards A,N side :

If the odd ball is A, it must be heavier than normal one. If the odd ball is C or D, it must be lighter than normal one.

Weigh C against D (3rd time):

If not balanced, the lighter one is the odd ball.

If balanced, A is the odd ball and heavier.

1a2/ If the scale tipped towards C,D side :

If the odd ball is A, it must be lighter than normal one. If the odd ball is C or D, it must be heavier than normal one.

Weigh C against D (3rd time) :

If not balanced, the heavier one is the odd ball.

If balanced, A is the odd bal and lighter.

1b/ If balanced (AN=CD) : B is the odd ball.

Weigh B against a normal ball to know if it is lighter or heavier than normal one .

*Last edited by tt (2005-07-04 11:33:46)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,654

Hi tt, and welcome to the forum

That is a good solution, and would look good on TV, I reckon.

Hero says "you are going to weigh each one? Harumph!" *throws four on one side, four on the other* *switches a few balls around* "switches one more time* "Here- this one!"

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**tt****Member**- Registered: 2005-07-03
- Posts: 32

It helps if you have skills with balls. (j/k)

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,654

Oh yes, I love billiards (ahem)

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**justlookingforthemoment****Moderator**- Registered: 2005-05-26
- Posts: 2,161

What does tt stand for, tt?

Or isn't it meant to stand for anything?

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**tt****Member**- Registered: 2005-07-03
- Posts: 32

It's short for my email address.

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**Roraborealis****Member**- Registered: 2005-03-17
- Posts: 1,594

Aha, I see. Heeheeheeheeheeheehee.......

School is practice for the future. Practice makes perfect. But - nobody's perfect, so why practice?

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**Zach****Member**- Registered: 2005-03-23
- Posts: 2,075

"tt" stands for "Too true", actually.

Boy let me tell you what:

I bet you didn't know it, but I'm a fiddle player too.

And if you'd care to take a dare, I'll make a bet with you.

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**rszopa****Member**- Registered: 2005-10-04
- Posts: 1

hello. i'm sorry to say it, but the neat answer isn't as new as you would expect... it has been published in at least five books (however, it isn't very popular in the Internet, if it makes you feel proud ). just take a look at http://www.cut-the-knot.com/blue/OddballProblem1.shtml. however, i'm not accusing anybody of plagiarism--this solution had been found many times independently (quite a good exapmle of convegence, you would say).

the problem with this solution is the rules of inference cannot be observed so clearly as in the ``classical'' solution. it would be quite good if someone (the author?) explained what is the rule of putting the balls on the left and on the right throughout the 3 weightings.

*Last edited by rszopa (2005-10-04 10:09:28)*

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,654

Mathematics is like that ... Pythagoras was not the first to discover the "Pythagoras" Theorem, for example!

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**wcy****Member**- Registered: 2005-08-04
- Posts: 117

who can juggle 12 balls?

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,654

I can, if they are all placed in a bag!

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**justlookingforthemoment****Moderator**- Registered: 2005-05-26
- Posts: 2,161

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**koundinya17****Member**- Registered: 2011-09-07
- Posts: 1

Instead of dividing it into 3 groups...why not into 4 groups of 3 each

(1) 3

(2)3

(3)3

(4) 3

1) (1) vs (2) - if equal weight

(1) vs (3) - from this attemp you will find which has the odd one and you will find in which group that odd one is

Now you got one group having 3 balls..in which the odd ball is

now 1 ball vs 1 ball ...solution found in 3 attempts

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi koundinya17;

Welcome to the forum! That is a solution!

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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