Weighing the balls

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Ýou have 12 balls.
All of them have the same mass.
Except one.
You don't know if it weighs more or less than the other 11 balls.
Your mission is to find the odd one out, and say if it weighs more or less than the 11 others.
You have a balancescale.
You may use it for three weighings.
The technique must work no mmater where the odd ball is.

EDIT: Added the balance part
 
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DaveC426913

Gold Member
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I think there's one item missing from the givens. Do you know the weight of the balls, or is that unknown?
 
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There is no need to know the masses of the balls in order to complete the brain teaser.

The correct answer will fill many A4:s of possible weighings.

ex.

If i weigh those balls against those, which one will weigh the most? How and in which weighingss do I get controlballs etc.
 
DaveC426913 said:
I think there's one item missing from the givens. Do you know the weight of the balls, or is that unknown?
I think the missing item could be that it's not a *scale* per se, so much as a balance. So, one weighing might be to have ball A vs. ball B, the result of which would tell you which was heavier, A or B. Doing that, then yes, it's possible within 3 weighs, and no other given information.

If it's really a *scale*... then... hm. Not sure how many weighs you'd need, but I'd bet a bit more than 3. I get between 3 and 6 weighs using a scale, 3 if you make lucky picks, 6 if you're unlucky. Better than 6, anyone?

DaveE
 
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Oh, you meant it that way! Yes it is a balance scale. You get the masses of the balls relative to each other and nothing else. Sorry for the confusion.
 
davee123 said:
If it's really a *scale*... then... hm. Not sure how many weighs you'd need, but I'd bet a bit more than 3. I get between 3 and 6 weighs using a scale, 3 if you make lucky picks, 6 if you're unlucky. Better than 6, anyone?
Came back to this and noticed it was possible in 5 (assuming a scale, not a balance). Considering that by effectively doing a binary-search-type weighing, you could do weighs of 6 balls, then 3, then 2, then 1, so maybe it's possible in 4 scale-weighs? But 5 might be the minimum considering that you need at least 4 weighs to *find* what you're looking for, plus 1 weigh to find out whether the one you're looking for weighs more or less than the remaining 12.

DaveE
 
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It is a balance.
 
Mattara said:
It is a balance.
But the scale weighing is much more intersting to me... The balance ones have been way overdone.
 
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The question is with a balance. Solve it. If you want a scale post another brain teaser
 
Mattara said:
The question is with a balance. Solve it. If you want a scale post another brain teaser
It can't be both?

The "find-the-1-of-12-objects-that-weighs-a-different-amount" problem has been done a bunch of times on this forum, so it's entirely possible that you won't see much of a response. The last time someone posted it, you can see the response they got:

https://www.physicsforums.com/showthread.php?t=105118

So, don't feel too bad if there isn't a reply.

I don't remember seeing a *scale* question before, though, so I know at least I was interested (and apparently Moo, too) in seeing a new problem. But, anyway, I'm done. Unless someone wants to discuss the scale problem further...

DaveE
 

NateTG

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Moo Of Doom said:
But the scale weighing is much more intersting to me... The balance ones have been way overdone.
If all you care about is worst case, then the scale is a pretty straightforward binary search.
 
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Ok, i didn't realize that it had been posted to death.

Ok it is a scale instead of a balance
 
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Mattara said:
Ok it is a scale instead of a balance
I have just worked it out in 3 steps using a plain balance so you don't really need a scale. (No, I didn't look up the solution.) It's a bit tricky, not just a binary search.
 
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ok, post the answer in the colour ;P
 
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Mattara said:
ok, post the answer in the colour ;P
Ok, I'll try COLOR="#e9e9e9" to match the "PF Prime" color scheme. I hope it works well enough with other schemes too...


Eleven balls have mass M, the target ball has a greater or lesser mass.

First, balance four balls against four balls.

Case 1: the balance is level so the target is in the remaining goup of 4. Label these four balls 1, 2, 3 and 4. Set aside ball 4 and balance group {1,2,3} against known balls {M,M,M} from the remaining eight.

Case 1.1: the balance is level so ball 4 is the target. A third balance determines if it is heavy or light.

Case 1.2: the balance tips towards {1,2,3} so one of these balls is the target, which is now known to be heavy. Balance ball 1 against 2 to find out which of these three is the heavy one.

Case 1.3: the balance tips towards {M,M,M} so one of {1,2,3} is the target, which is now known to be light. Balance ball 1 against 2 to find out which of these three is the light one.


Case 2: the first balance was not level so the target is in this group of 8 balls. Label balls on the heavier side {1,2,3,4} and balls on the lighter side {5,6,7,8}.

Now balance balls {1,2,M} against balls {3,4,5}.

Case 2.1: the balance is level so the target is in remaining set {6,7,8} and is known to be light (from the first balance). Balance 6 against 7 to know which of these three is the light one.

Case 2.2: the balance tips towards {1,2,M} so either a ball in {1,2} is heavy or ball 5 is light. Balance ball 1 againt ball 2 to know if one is heavy and which one, or to know that ball 5 is light if the balance is level.

Case 2.3: the balance tips towards {3,4,5} so either ball 3 or 4 is heavy. Balance 3 against 4 to know which one.


I hope I didn't screw up anywhere. :uhh: Someome will let me know for sure.
 
i propose a solution shud be right ...enjoy

i will use W for the word weigh..u remove 2 coins .so u W the 2 piles of 5 coins .if they r the same then the 2 coins contain 1 odd coin so u take them and wiegh them against each other now u no 1 is light or normal or the other is heavy or normal, so now u choose any of them and weigh them against a normal coin.

if the piles of 5 r difirent u name 1 pile heavy pile and 1 pile light similar to wat we did with the 2 coins. we remove a light coin put it aside and we form groups HHL HHL HLL we weigh HHL vs HHL if they weigh diffirent then odd coin is either H ,H or L , ( call normal coin N) so u do HL vs NN leaving out L .. if HHL vs HHL shows they weigh same then odd coin in HLL with 1 wiehing left u do HL vs NN leaving out L ...QED

tell me if u find me incorect.(btw i didnt write all details in last part u can easily fill in the gaps)
 
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