Finally, here is the solution to this enigma... I hope you enjoyed it, and that you'll enjoy the one of this week...
I wanted to make a drawing because I thought it would be clearer but it was taking me too much time, and it wasn't that clearer, so here is the solution explained :
YOu start by dividing your coins in three groups of four coins. Let's call these groups A, B, and C. Here below is the notation we'll take :
A is composed of A1, A2, A3 and A4
B is composed of B1, B2, B3 and B4
C is composed of C1, C2, C3 and C4
First weighing :
We'll compare A and B. It means A1, A2, A3 and A4 one side and B1, B2, B3 and B4 the other side.
The result of this weighing can only be either a stability either an instability.
In the case of a stability, it means the fake coins belong to the C group.
Second weighing :
we compare C1 + C2 one side and C3 + A1 the other side.
We use A1 because we know as the result of the first weighing that it isn't a fake coin. We could have chosen any other coin of A or B.
The result of this weighing can be :
- stability : It means the fake coin is the one we haven't touched : C4
- C1 + C2 is heavier than C3 + A1 : It means that either C1 or C2 is the fake coin and it's heavier, either C3 is the fake coin and it's lighter (because A1 is known to be a real coin) (situation *)
- C1 + C2 is lighter than C3 + A1 : It means that either C1 or C2 is the fake coin and it's lighter, either C3 is the fake coin and it's heavier. (situation **)
In both cases :
Third weighing :
We compare C1 one side and C2 the other side.
- Stability : The fake coin is C3
- situation * : The fake coin is the heaviest of this weighing
- situation ** : The fake coin is the lightest of this weighing
Let's return to the first weighing know. What happens if we have an instability ?
Well, we know that the fake coin belongs to the 8 that we have just weighted. It means to A or B. Then it's a little bit more difficult...
We make two new groups:
D composed of A1, B1 and B2
E composed of A2, B3 and C1
TO make it easier to understand, let's consider that at the first weighing, A was heavier than B. It's just a matter of notation.
Second weighing :
We compare D one side and E the other side.
If we have stability, it means the instability we had at our first weighing was due to one of the coins we have taken away. (A3, A4 or B4)
Third weighing :
We compare A3 one side and A4 the other side.
If we have stability, the fake coin is B4 and it was lighter (because of the first weighing)
If not, the fake coin is the heaviest between A3 and A4 because we know thanks to the first weighing that A is heavier than B. As the fake coin is between two coins from A, it can only be heavier.
If we don't have stability at the second weighing :
If D is heavier than E :
The fake coin is either A1 either B3 (thanks to the first weighing that told us that A was heavier than B. Which means that the fake coin is either an heavier one from A either a lighter one from B)
Third weighing :
We compare A1 one side and C1 the other side. (we chose C1 because we're sure it isn't fake)
If stability, it means the fake coin is B3 and it's lighter.
If not, it means the fake coin is A1 and it's heavier.
Let's turn on to the last situation : D is lighter than E.
Then it means B1 or B2 is the fake coin and it's lighter or A3 is the fake coin and it's heavier.
Third weighing :
we compare B1 one side and B2 the other side.
If stability, it means the fake coin is A3 and it is heavier.
If not, the fake coin is the lightest between B1 and B2 (thanks to the first weighing).
Finally we got all the situations possible !!!
This enigma wasn't easy at all, congratulations, to Florian who managed to find the solution !!
