# please solve this riddle



## sham1983 (Dec 18, 2008)

We have total 12 balls in which 11 balls have same weight but only ball has different weight. You have allow three times to weight all balls and remove one ball which has different weight and tell me the weight of that ball.


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## Domski (Dec 18, 2008)

Maybe:

Split the balls into 3 sets of 4 balls. Weigh 2 sets of 4 and find out if equal or not. From that you will know if one of the sets of balls weighed is lighter or if they are the same, if the same then the lighter set must be the set that you haven't weighed.

Split the lighter set of balls into sets of 2 balls. Weigh again to find the lighter set of 2 and then weigh each of the balls from the ligher set to determine which is the lightest ball.

Dom


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## SydneyGeek (Dec 18, 2008)

Similar:

Weigh 6 on each side, take the lighter set and split to two sets of 3.
Now weigh any two of the lighter triplet. If they balance, the lighter one is the one you didn't weigh in the third go. If they don't you will know which is the lightest. 

Denis


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## Andrew Fergus (Dec 18, 2008)

This took me a wee while to work out but if you don't know if the different ball is heavier or lighter then this should do it:

Weigh 1 to 4 vs 5 to 8.

If 1 to 4 = 5 to 8, then weigh 1 & 9 vs 10 & 11.  If these are the same weigh 1 vs 12, otherwise 1 & 12 vs 9 & 11.  Different outcomes give varying answers as to which ball is different and whether it is lighter or heavier.

If 1 to 4 < 5 to 8, then weigh 1, 2 & 5 vs 3, 4 & 7.  If these are the same weigh 10 vs 8, but if 1+2+5 < 3+4+7 then weigh 1 vs 2, else 3 vs 4.  Again, different outcomes for the 3rd weigh give different answers as to which ball is different and whether it is lighter or heavier.

See if you can work out the 3rd sequence if 1 to 4 > 5 to 8.

Andrew

P.S. Done properly, you should be able to cover all 24 instances of balls 1 to 12 being either heavier or lighter.


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## Stormseed (Dec 18, 2008)

Hello,

As the OP says, that you are allowed to weigh the balls for 3 times and get the weight of the 12th ball which infact is a lighter ball. We now have an *idea* as how to go about finding out the lighter ball.

How do we arrive at a conclusion in respect to the *weight* of the ball ? Anyone having some pointers ?


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## schielrn (Dec 18, 2008)

Stormseed said:


> How do we arrive at a conclusion in respect to the *weight* of the ball ? Anyone having some pointers ?


I would assume if we are weighing the balls, we are using some sort of scale to get weight.  When I first saw this, I thought of Domski's approach.


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## cornflakegirl (Dec 18, 2008)

But all the approaches rely on old-fashioned scales - measuring one (set of) ball(s) against another - so no actual weights involved.


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## MrKowz (Dec 18, 2008)

If it is a scale that gives an actual weight instead of a scale that compares weight....

```
1) Weigh a random ball
2) Weigh a second random ball
    If (1) and (2) weigh the same, then you know that is the weight of the heavier ball.
    If (1) or (2) is lighter than the other, then the lighter one is the weight of the lighter ball.
3) Weigh ALL balls, the total weight = 11*(weight of heavier ball) + weight of lighter ball.
```


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## Stormseed (Dec 18, 2008)

How about this ?


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## markelly (Dec 18, 2008)

OK, but the question is about the weight of the lighter ball as well.
I don't think you can do it with three steps only.


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## Jonmo1 (Dec 18, 2008)

If you mean weight by using a scale, you can get the wight of the lightest ball by using a scale only once. All other steps involve COMPARING weight by just determining which is heavier, not needing to know how much each weighs...

Doms approach is the best.

4 sets of 3.

Put balls 1-4 in a box
put balls 5-8 in a box

Lift each one and an average person should be able to tell which one is lighter, or if they are the same. You have not used a scale.

If the same, then the lighter ball is in the group 9-12, otherwise, the lighter ball is in either group 1-4 or 5-9.

Take the group with the lighter ball in it, put 2 balls in 1 box, the other to in another box.

Pick them up and determine which is lighter. Still have not used a scale.

Then pick up each ball of the lighter group of 2, and you can tell which is lighter.

Then use the scale for that lightest ball to get it's weight.


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## Andrew Fergus (Dec 18, 2008)

Unless I'm misunderstanding something, where in the question did it say the odd ball was lighter?


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## klb (Dec 18, 2008)

Note that the question says you can only weigh all balls 3 times and it does not say that you can only use the scale 3 times.  My preference is to use a digital scale - not a balance scale.

Divide balls into 3 groups of 4
Weigh each group.  
One group will have a different weight - either higher or lower than the other two.
Weigh each ball in the odd group.  Read the weight on the scale to determine the weight for each and you now know the weight of the ball which is different.

You have weighed 8 of the balls only once and the other four a maximum of twice.


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## Jonmo1 (Dec 18, 2008)

> Unless I'm misunderstanding something, where in the question did it say the odd ball was lighter?


 
Lighter / Heavier, doesn't matter.  The same logic applies.



> Note that the question says you can only weigh all balls 3 times and it does not say that you can only use the scale 3 times


If you can weigh EACH BALL 3 times, then it's not a very good riddle.

You can just weigh each one a single time (one at a time) to dermine which ball is different...


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## klb (Dec 18, 2008)

The riddle lies in how it is interpreted or misinterpreted.  Then again it is not my riddle and not likely to happen in real life.

I like your idea of weighing each ball.  Keep it simple.  Potentially you only use the scale three times if one of the first three is the odd ball - lol  
With my "simple" solution you use the scale a max of seven times and none of the balls are weighed three times.


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## Andrew Fergus (Dec 18, 2008)

jonmo1 said:


> Lighter / Heavier, doesn't matter.  The same logic applies.


Yes it does.  If you weigh the lighter set of 4 balls a 2nd time, then you have wasted one of the maximum of three weighs if the different ball was heavier.  That's the point - you don't know if it is heavier or lighter.  If a relative scale with two set of four balls tips to the left, was one of the balls on the left side lighter, or one on the right side heavier?  You cannot tell with a single weigh - by pre-measuring relative weights of balls in boxes, you have used one (or more) of your maximum number of weighs.

I believe the OP has been a little unclear in stating the problem.  I found the original problem stated correctly here : http://256.com/gray/teasers/



 There are 12 balls.
 11 of the balls weigh the same.
 1 of the balls is _either_ heavier or lighter than the rest.
 You have an unmarked balance scale.
 Using the scale only 3 times, determine which ball is different and whether it is heavier or lighter than the rest.
I think you'll find the solution I worked out provides the correct answer.

Andrew


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## cornflakegirl (Dec 19, 2008)

I did find an interesting extension to the problem: can you devise a solution whereby you weigh the same sets of balls each time, regardless of the results you obtain?


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