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Originally Posted by Meorazhar
I have solved this one  Can't say it was easy though. Anyone else? |
I think the balance would be used to eliminate the heavier coins.
1st split 6 and 6, eliminate the heavier stack, the stack that is eliminated will always hold all the same coins.
2nd split 3 and 3, again eliminate the heavier stack.
3rd, only place 1 coin on one side and 1 coin on the other side of the balance.
If the last time using the balance was the same then the coin that wasn't placed on the balance would be the correct coin. If it was the other way around, the coin on the left or right of the balance from the 3rd turn would give the answer.