# Best Puzzle I ever Heard



## SparkRite (26 Mar 2010)

Ok folks, just to lighten the atmosphere here on AAM.
No trick wording or question, simply logic..........

There are 12 coins, each identical in appearence.
One is of odd weight.
You have a balance.
You are allowed three uses of the balance.

Find the odd coin and also whether it is lighter or heavier that the remaining 11.



I remember "Time" magazine offer $1000 to the first correct answer about 20 years ago or so.

I know the answer, myself and my maths lecturer worked it out, but it took some time.


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## Caveat (27 Mar 2010)

I've a feeling equations might come into it and if so, I'm out.


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## mathepac (27 Mar 2010)

I was dreaming about this bleedin' thing last night, doing binary splits and other fancy dance routines. I'm knackered and it's only 9:25am.


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## z107 (27 Mar 2010)

If by balance, you mean something that looks like this: --/\--
(Rotten solution deleted)

(If a grand was on offer, I'd brute force the combinations to find the solution. Shouldn't take that long considering number of coins on each side of the scale should be equal)


+++ Update - think I have it now +++
Step One
Divide into four piles of three, A B C and D
Weigh A against B
Weigh B against C

If A, B and C are all the same, D should go to next step
If A is different to both B and C, note if it's lighter or heavier - Use A for next step (Do same for B and C)

Step two - three coins left from step one
Coins E, F and G
Weigh E against F, if the same, odd coin is G, otherwise it's the lighter or heaver (see above) or either E or F.


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## ninsaga (27 Mar 2010)

I would just drop all coins together from the same height & see which one either hits the ground first or last.


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## Welfarite (27 Mar 2010)

ninsaga said:


> I would just drop all coins together from the same height & see which one either hits the ground first or last.


 It wouldn't matter, they'd all hit the ground at the same time regardless of weight!


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## Capt. Beaky (27 Mar 2010)

Welfarite said:


> It wouldn't matter, they'd all hit the ground at the same time regardless of weight!


How does this apply? If one of the coins was made of polystyrene and the rest of lead or vice versa


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## z107 (27 Mar 2010)

I'd buy another balance with one of the coins


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## z107 (27 Mar 2010)

> How does this apply? If one of the coins was made of polystyrene and the rest of lead or vice versa


That's air resistance.


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## SparkRite (27 Mar 2010)

umop3p!sdn said:


> If by balance, you mean something that looks like this: --/\--
> (Rotten solution deleted)
> 
> (If a grand was on offer, I'd brute force the combinations to find the solution. Shouldn't take that long considering number of coins on each side of the scale should be equal)
> ...



Hi umop3p!sdn, I see you have put a bit of thought into that,but unfortunatly you are not correct, you have not covered ALL the variables.

BTW. You are correct, by balance I do mean this  --/\--


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## Crugers (28 Mar 2010)

Make these three weighing with 4 coins on each side. The result will be either BB(BothBalance) or UD(UpleftandDownright) or DU(DownleftandUpright)
1,2,3,4 V's 6,7,9,11 The result will be one of BB/UD/DU
2,5,8,11 V's 3,4,6,10 The result will be one of BB/UD/DU
5,4,11,12 V's 1,6,7,8 The result will be one of BB/UD/DU

Then refer to the list below..
Coin #1 Heavy: BB,UD,DU.
Coin #1 Light: BB,DU,UD.
Coin #2 Heavy: DU,BB,DU.
Coin #2 Light: UD,BB,UD.
Coin #3 Heavy: UD,BB,DU.
Coin #3 Light: DU,BB,UD.
Coin #4 Heavy: UD,DU,DU.
Coin #4 Light: DU,UD,UD.
Coin #5 Heavy: DU,DU,BB.
Coin #5 Light: UD,UD,BB.
Coin #6 Heavy: UD,UD,UD.
Coin #6 Light: DU,DU,DU.
Coin #7 Heavy: BB,UD,UD.
Coin #7 Light: BB,DU,DU.
Coin #8 Heavy: DU,UD,BB.
Coin #8 Light: UD,DU,BB.
Coin #9 Heavy: BB,BB,UD.
Coin #9 Light: BB,BB,DU.
Coin #10 Heavy: UD,BB,BB.
Coin #10 Light: DU,BB,BB.
Coin #11 Heavy: DU,DU,UD.
Coin #11 Light: UD,UD,DU.
Coin #12 Heavy: BB,DU,BB.
Coin #12 Light: BB,UD,BB.
So print off the list and carry it around with you in case you ever find yourself in that particular situation!


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## truthseeker (29 Mar 2010)

...


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## homeowner (29 Mar 2010)

Put 6 coins on each side of scales, which ever side is heaviest contains the heavy coin.

Discard the other 6.

Put 3 coins on each side of scales, which ever side is heaviest contains the heavy coin.

Discard the other 3.

From the 3 remaining coins, place 1 on each side of scales.  You will either see one side being heavier (ie thats the heavy coin) or both will be equal weight, in which case the heavy coin is the one not on the scales.


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## Welfarite (29 Mar 2010)

homeowner said:


> Put 6 coins on each side of scales, which ever side is heaviest contains the heavy coin.
> 
> Discard the other 6.
> 
> ...


 Brilliant except for one thing; you presume the coin is heavier; it may be lighter. The quest is to determine the 'odd' weighted coin and whether it is lighter of heavier than the others in 3 uses of the scales.


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## mathepac (29 Mar 2010)

homeowner said:


> Put 6 coins on each side of scales, which ever side is heaviest contains the heavy coin...


This was the start of one of my attempted solutions buy it can't work because - 


SparkRite said:


> ... Find the odd coin and also *whether it is  lighter or heavier* that the remaining 11...


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## homeowner (29 Mar 2010)

Welfarite said:


> Brilliant except for one thing; you presume the coin is heavier; it may be lighter. The quest is to determine the 'odd' weighted coin and whether it is lighter of heavier than the others in 3 uses of the scales.



Oops, i misread the question.....


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## z107 (29 Mar 2010)

Is Crugers solution correct?


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## DB74 (29 Mar 2010)

I would say that Cruger's solution is probably the correct way of thinking although the solution actually posted is incorrect:

If Coin 1 is heavier or lighter then the first "weighing" would not balance
If Coin 12 is heavier or lighter then the last "weighing" would not balance


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## SparkRite (29 Mar 2010)

Cruger is not correct and is overly complex...........

At least I don't think it is, did my head in trying to follow it.....


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## Crugers (29 Mar 2010)

Crugers said:


> Make these three weighing with 4 coins on each side. The result will be either BB(BothBalance) or UD(UpleftandDownright) or DU(DownleftandUpright)
> 2,5,8,11 V's 3,4,6,10 The result will be one of BB/UD/DU
> 5,4,11,12 V's 1,6,7,8 The result will be one of BB/UD/DU
> 1,2,3,4 V's 6,7,9,11 The result will be one of BB/UD/DU
> ...


 
Apologies... Copy and paste from excel gave terrible visual results so I had to cut and paste piece by piece..
Got the order of the three weighings, out of order...
I think I've got it in the correct order this time...
(until someone tells me different... )


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## Howitzer (29 Mar 2010)

That's the answer but not the solution. 

How do you determine which coins to weigh against which? 11,6 are weighed 3 times, 9,10 just once.


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## DB74 (29 Mar 2010)

OK - how 'bout this

1. Weigh up coins (1-3) vs (4-6) - if they don't balance then go to step 2, if they do balance go to step 4

2. If the first 6 don't balance then select 3 new coins (7-9) and weigh up against 3 of the original 6 (1-3) - this will let you know if the odd-one out is in group (1-3) or (4-6) and also whether the coin is heavier or lighter

3. Once you know which group of 3 the coin is in you just take 2 of those 3 coins and weigh them up - if they balance it's the other one - if not you already know if the coin is heavier or lighter so it's obvious which it is.

4. If they balance, then the odd coin is in the group 7-12.

5. If 1-3 vs 7-9 doesn't balance then you know if the coin is heavier or lighter so select 7&8 and if they balance then 9 is the odd-one-out and if not then you know if it's 7 or 8

6. If 1-3 vs 7-9 balances then you know that the odd-one-out is 10-12 but you only have one weigh left so *you're screwed!*

*AAAAAARRRRRRGGGGGGHHHHHH!!!!!!!!!!*


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## SparkRite (29 Mar 2010)

Cruger, that seems to work out all right, but is not really practical.

How the hell did you work out which coins to weigh against which other coins??

Anybody want a little clue?


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## orka (29 Mar 2010)

Weigh 4 coins against 4 coins. 

A: If they balance, the odd one is one of the remaining 4. Weigh 3 of the remaining 4 against 3 known normal ones. If this balances, the only remaining unweighed coin is odd and can be tested against a normal to see if it is lighter or heavier. If this doesn’t balance, the odd is in the 3 recently weighed – and you’ll know if it heavier or lighter because the 3 are being weighed against normals. Weigh 2 of the 3 against each other to work out which of the three is odd.

B: If they don’t balance, remove three coins from the heavy side, replace with three coins from the light side. Put three known normals on the light side.
If thy balance, one of the 3 heavy side coins removed is a heavy coin – weigh two of them against each other to figure out which one.
If the side that was heavy is now light, one of the 3 coins moved from the light side to the heavy side is a light coin – weigh two of them to figure out which of the three. 
If the heavy side stays heavy, EITHER the unmoved heavy side coin is heavy or the unmoved light side coin is light – test one of these against a known normal to find out which.


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## SparkRite (29 Mar 2010)

orka said:


> Weigh 4 coins against 4 coins.
> 
> A: If they balance, the odd one is one of the remaining 4. Weigh 3 of the remaining 4 against 3 known normal ones. If this balances, the only remaining unweighed coin is odd and can be tested against a normal to see if it is lighter or heavier. If this doesn’t balance, the odd is in the 3 recently weighed – and you’ll know if it heavier or lighter because the 3 are being weighed against normals. Weigh 2 of the 3 against each other to work out which of the three is odd.
> 
> ...



EXCELLENT, well done Orka.

The secret is in the second use of the balance, by transferring three from one side to the other,which is in fact giving you information about at least 9 coins and often also whether the odd one is heavier or lighter.

The easiest scenario is when the first use balances.


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## Staples (30 Mar 2010)

This is one of the puzzles conatined in the "Professor Layton and the Curious Village" game.  

If you like this type of puzzle, you'll love this game- and its successor "Pandora's Box".


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## DB74 (30 Mar 2010)

I was actually going to buy Pandora's Box for the DS last weekend but the puzzles featured on the front of the packaging looked very childish so I didn't bother in the end.


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