Robert Hansen posted Mar 8, 2013 12:39 AM (GSC's remarks interspersed): > > lol, I really didn't think my answer would have > stumped you. > lol, your answer stumped me - and I had a short time ago posted a question expressing my bewilderment. > >Here is a simpler version. Split the > group into two groups of 6, keep the heavier group, > split that into two groups of 3 keep the heavier > group, weight (any) 2 coins from the remain group of > 3, if they weigh the same then the other coin is the > hevier one, if not then the heavier one is the > heavier one. > > It is a binary search. > > Bob Hansen > Your 'simpler version' is just plain and simply wrong. Read the puzzle more carefully.