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littleprince, it's not quite as easy as that since he did say that one of the coins is lighter or heavier than the others.
[edit]: oops, didn't see Haircut's message...
but it applies to what IJump said too. >>
Then if on the second weigh, you find they are even, weigh the other group. The other group has the conterfeit. That would add two to the overall. I missed the "or heavier" part of the original problem.
So that would make the total 5 at the most......... hmm...