I just came across your forum and the weighting of 12 coins. I was wondering if I can present you with a posible solution ?.

lets divide the coins in 4 groups of 3:

A has 1,2,3,
B has 4,5,6
C has 7,8,9
D has 10,11,12

With Givens as 11 of the coins have the same weight..

By dividing this into 4 groups We know that 3 out of 4 carry the same weight. Next, group A and B on one side and C and D on the other. Record the way the balance tips. ( First weighting). Lets just assume that A/B is heavier. Then We swap position of A and D. ( 2nd weighting). Anh lets assume that D/B is heavier this time around. From this, and the givens that 3 out of 4 groups
are the same, We can determine that B is the heavier group, hence the group that contains the counterfeit coin.

>From group B we take any 2 coins and scale them. If they are the same, in weight, The last coin is counterfeit. If they are not. The heavier is the counterfeit one. (Now this because I assume A/B was heavier, if A/B was lighter, the lighter coin is the couterfeit)

This could help with the help of algebra:

1... A + B >< C + D. (first weigting)
2... D + B >< C + A (second weighting)

add (1) and (2)

We have 2B><2C which is B><C

givens was that 11 coins are the same , so at least we have the 3 groups which are the same. with B><C We can safely say that A=D. hence the coins in the two groups are the same.

We now have B><C with either one of these two conditions A = D = B>< C ----- or B>< C = A = D.

Thanks for reading, and if you can share with me your thoughts on this.

Thank you

Thomas Le.