# Thread: Counterfeit coin in three weighings

1. ## Counterfeit coin in three weighings

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 balance AND determining 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 the third weighing against 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.
The third weighing of 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.
The third weighing of 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.

2. cthree posted a variation of this puzzle some time ago. You can also read more about it here. It is a challenging puzzle.

3. Thanks, gabi. I had done a search on counterfeit coins, but of course did not find that thread as it is about statues. Your solution is different from mine and I'll have to have a closer look at it after I have had a good night's sleep.

4. Molnar has a more detailed explanation than mine in post #9 of the statues' thread.

5. Originally Posted by gabi
Molnar has a more detailed explanation than mine in post #9 of the statues' thread.
I have read Molnar's explanation and the first section is different from mine, but achieves the same results. Molnar's method there is more efficient.

However, in the next part I find a flaw in the argument where it says "The results from the first weighing show if the odd statue is heavy or light". The first weighing cannot establish this. The side that is lighter than the other in the first weighing might be so because if contains a light statue or because the other side has a heavy statue.

I have yet to work through your explanation. I'm not awake enough for that yet - just give me a few more hours.

6. Originally Posted by knocka
Originally Posted by gabi
Molnar has a more detailed explanation than mine in post #9 of the statues' thread.
I have read Molnar's explanation and the first section is different from mine, but achieves the same results. Molnar's method there is more efficient.

However, in the next part I find a flaw in the argument where it says "The results from the first weighing show if the odd statue is heavy or light". The first weighing cannot establish this. The side that is lighter than the other in the first weighing might be so because if contains a light statue or because the other side has a heavy statue.

I have yet to work through your explanation. I'm not awake enough for that yet - just give me a few more hours.
Now that I'm wide awake I have worked though Molnar's explanation. The results from the first weighing per se do not show if the odd statue is heavy or light, but the results do show for subsequent weighings whether the odd one is potentially heavy or light and that is how the results were used, as I found out. I should have worked through the whole method instead of interpreting one sentence literally.

Molnar's method is far more efficient than mine, but I feel pleased that I did work through and find a method that works too. I'll leave my method on my TweSpace.

Gabi's method lists most of the permutations possible with three variables, U D M. MMM is not listed as that would only occur if there were no counterfeits. The other two not listed are DUU and UDD which both result in contradictions.

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