This is not an original puzzle, but it was new to me.

In Chat bob1946 raised a problem about detecting one counterfeit coin in a group of 12 coins with just three weighings on a spring balanceANDdetermining whether the counterfeit was lighter or heavier than a genuine coin. I had met this type of problem before when the information about whether the coin was heavier or lighter was known in advance. With the twist of not knowing about the weight of the counterfeit the problem became more complex and needed quite some thought before the AHA moment finally came.

I have written up most of the answer on a page in my TweSpace. It is not complete, but the last part is just a repetition of what has already been written with the swapping of the words light and heavy.

There may be others like me for whom this problem is new and they may want to think about working out a solution.

First Part of solution: coins 1,2,3,4,5,6,7,8,9,10,11,12.

WEIGHING #1.

(1,2,3,4)/\(5,6,7,

If (1,2,3,4)=(5,6,7,, then 1,2,3,4,5,6,7,8 are all genuine and the counterfeit is one of (9,10,11,12) and could be light or heavy.

WEIGHING #2.

(9,10,11)/\(1,2,3)

If (9,10,11)=(1,2,3), then the counterfeit is 12 and thethird weighingagainst any genuine coin will determine whether it is light or heavy.

If (9,10,11)<(1,2,3), then the counterfeit is one of (9,10,11) and is light.

Thethird weighingof 9/\10 will determine whether it is 9 or 10 which is the light one or, if 9=10, then 11 is the light counterfeit coin.

If (9,10,11)>(1,2,3), then the counterfeit is one of (9,10,11) and is heavy.

Thethird weighingof 9/\10 will determine whether it is 9 or 10 which is the heavy one or, if 9=10, then 11 is the heavy counterfeit coin.

If (1,2,3,4) does not equal (5,6,7,, then there is a more complicated procedure to follow. I'll not post it here yet in case others want to work through this puzzle.

## Bookmarks