Can you find the coin?

[KangaRoo]

This is a riddle about 12 coins. When you look at them they appear identical. But one of them is different. It's got a different weight to the other 11. It may be heavier then the other 11 coins, or, it may be lighter then the other 11 coins. That is what you have to find out.
You are given a set of balancing scales. You may place any number of coins on either side of the scales. For example, 6 on one side and 6 on the other side of the balancing scales. That would constitute one weighing cycle. You are only allowed 3 weighing cycles to find the coin that is different.

Good luck, you got 30 minutes to solve this riddle. ;)

[Meatmallet]

this may work
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this took me as long as it took to type so it may be wrong

and I done this sober so it may never work :lol:

[danariba]

Lets take a look

[KangaRoo]

Hi MM, it's not as easy as that m8, in your example 5 and 5 = equal, then the other two will not be equal but which one is the odd one? The odd one may be heavy or light, and you have not told me which one.
If 5 and 5 not equal then 4 and 4 does not work either. Don't forget it's not only the odd coin we are looking for. You have to determine if the coin is havy or light as well.
Sorry mate, but no cigar for that effort. :D

[Cadd50]

I'll give it a go, I think.

[copytech]

i'll try it

[Dude]

on my first try I got it down to where I could tell if the first 10 coins were off and if they were heavier or lighter, but if it was one of the last 2 coins then I need to weigh them a 4th time to tell. :(

so I am 10/12ths of the way to the right answer. :D

[KangaRoo]

You're doing well Elmer, 10/12ths is almost there, but it's the 2/12ths that's really hard.

[n141311]

NOTE right, i've assumed that 3 weighing cycles means: i can only put a set of coins onto the weigh no more than 3 times, but once a set of coins are on the balance i can play around with taking some off and on.

How i'd do it:

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I hope i'm right :)

[KangaRoo]

Hi there n141311, the riddle is not solved if you can tell me the 'dodgy' coin. You must be able to also say whether it is lighter or heavier then each of the other coins.
One weighing cycle is completed after placement of x number of coins on each side of the scales. You can take some or all the coins off the scale and substitute them with other coins, but that would constitute 2 cycles.

[BongoKongo]

Start off as you said with 6 on each side then you take the pile of 6 that is lighter and seperate them into 2 piles of 3 and weigh them. Then you take the set of 3 coins that are the lightest and and weigh 2 coins. If the 2 coins ballance then it is the third coin that you haven't weighed yet. If one of the 2 coins you are weighing is heavier than the other then you have found your light coin.
Simple as E=MC2
PS: The coin has to be lighter because when you weigh the 12 coins the pile of 6 coins that is the lightest has to have the different coin.
Reason:If you took the 6 coins from the heaviest pile and seperated them into 2 piles of 3 then these 2 piles of 3 would both have to weigh the same.
Who said too much pot when you are young numbs the brain???

[n141311]

hey kanga, my method would tell you which coin was the odd one...because if it was a different weight to any of the other coins then the scales would not balance.

Surely, my method is correct?

If not, then plleeeeeeeeeease PM your answer. I am intrigued as to how there could be a better answer than mine. :P

(yes, I am arrogant).lol

BONGO:

The thing with your method is that Kangaroo has NOT mentioned whether the odd coin is heavier or lighter. So, hypothetically if the odd coin was actually heavier and not lighter (as you assume in your answer) then you would inevitably come up with the wrong answer. The only way using your approach would be to do it twice using each assumption (that the odd coin is heavier / lighter)..but that would exceed the 3 x limit on weighing.

[Dude]

n141311, I like your assumption that you can take coins off and just count it as being weighed one time, I didn't think of that. :)
but in that case, you can find the coin in just 2 times.

- put 6 on each side and take them off one at a time until the scale balances. (that's one)

- take the last two coins you removed and weigh one against another coin and then you have your answer. (that's two)

anyway, I figured out the answer to this one, but it is one of the hardest riddles I think I have ever seen. and it sounds so simple, just weigh 12 coins, how hard could that be?

I will wait to post the answer to see what others come up with.

[BongoKongo]

Ok,point taken, then this does prove that smoking Pot when you are young does numb your brain.

[KangaRoo]

Hi n141311, I'm sorry m8, but your wrong. Using your method of setting up the scales with a set number of coins and then removing them one at a time is not allowed. That way a 5 year old could solve the riddle, as Elmer quite correctly pointed, out in 2 weighing cycles.
You have to do better then that m8 before I give you the solution. :D

[KangaRoo]

Hi Elmer, if you got the solution m8, please pm it to me so I can see your method of solving the riddle.

[KangaRoo]

Ok,point taken, then this does prove that smoking Pot when you are young does numb your brain.

Not neccessarily m8, you just need to approach the problem logically and analyse the criteria set. Here is a tip, number your coins from 1 to 12. It's a lot easier to follow that way when you move them around. :D

[Dude]

I solved the riddle,
but I am not exactly sure I can explain it so that anyone else could really understand it.
it gets kind of complicated.

I will give everyone a hint, because I wasted a lot of time when I started out... I was thinking that you should put 3 coins on each site to start, because that would split the coins into two groups of 6. and you would then know which half the odd coin was in.
But after a very long time I figured out if the first 6 are balanced then you only have 2 turns left and the odd coin is one of 6, and you can't find it in 2 turns with 6 coins.

But here is the really tricky part, if you put 4 coins on each side then if they are off, you have 2 turns left and you know it is one of the 8 coins and it is possible to find the odd coin, which logically does not seem to make sense in my mind but it works out that way.

anyway, I will work on the rest of it which gets a lot harder from that point on, and see if I can explain it without writing a book.

[KangaRoo]

Yep, your on the right track Elmer, 4 and 4 IS the starting point!