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Originally Posted by Alan Anderson
The details are simple but hard to explain without graphics. Label the twelve coins A through I. Weigh ABC against DEF, then ABC against GHI. The results will tell you which group of three is different and whether it is heavy or light. Then you weigh any two of the different group's coins against each other to find out which is the odd one; if they match, the third one is the different one.
if ABC < DEF and ABC < GHI, one of ABC is light
if ABC < DEF and ABC = GHI, one of DEF is heavy
if ABC = DEF and ABC < GHI, one of GHI is heavy
if ABC = DEF and ABC > GHI, one of GHI is light
if ABC > DEF and ABC = GHI, one of DEF is light
if ABC > DEF and ABC > GHI, one of ABC is heavy
The other three combinations are not possible if exactly one coin is different.
Once it's narrowed down to three coins, just compare two of them. If they don't balance, you already know whether the fake is heavy or light; just choose that one. If they do balance, the other must be the fake.
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I hate to break it to you after all your work, but there are 12 coins, and lettering them A through I is only 9.
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If I take two, then I can give one (half of my two) to each guard and according to the agreement he'll give it back to me, leaving me with the two I took.
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Yep you got that one!
Keep working on the 12 coin problem - its a real toughie!