# Solution to 12 coins balance problem

Hi,
in the last post I described this cool balance problem with 12 coins where one of them is false one (heavier or lighter than others). You are allowed to weight them 3 times to find out which one of them it is and if it lighter or heavier.

The hint that I gave you is to separate them on 3 groups of 4 coins. Otherwise it is not possible to solve it. For the first one there are two options as you weight 4 coins against 4 coins.

If the outcom is even it goes as follow:

1. If nothing happens the 4 coins that you have not used must have the false one. Now this is quite tricky. You have probably come out to this part but you have to use all the facts that one weighting will get you as you will see. If something goes down, it is good to know it and vica versa.
1. a) Now you know that the rest of the coins (4) have to have the false one since it was even. You take one away and put the three against 3 that you are sure about (you have 8 of those). If it is even you know that it is the last coin and you just weight it against one of those normal ones to see if it is lighet or heavier. If you find out that one of the three has to be false you will remember if they were heavier or ligher. Then you take two of them and weight them against each other. The one that did the same thing like in the last weighting is the false one. (If in the 3 vs 3 it was heavier you are searching for the heavier side again.) If they are even you know that it is the last one.
2. Now it can also happen that one side from the first weighting is heavier than the other. All of the 8 coins are unknown but you can use the fact that one went down and the other up so you need to remember that fact. You also need the 4 coins that you know must be all ok. Put three of those on the “lighter” part. From the lighter part move three original to “heavier” part and from there away three.
1. b) The easiest thing is when the heavier part is again heavier. Then it has to be one of the original coins from both side which are only 2. You weight one of them against any other coin to get the result.
2. b) If the part that was heavy is lighter this time, one of the three coins that you moved from lighter to heavier last time must be the wrong one. It is also lighter as you know. Weight two of them against each other. If they balance it is the third one if they do not it is the lighter part.
3. b) If they balance it must be one of the three coins that we removed completely. We use the procedure described in the last post to find the solution, it is heavy coin then.

Now that is for solution. It is not easy to come up with it and there are some even more difficult versions. Thanks for reading and thanks to Wikipedia for the solution.

Dragallur