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.

You must be mistaken somewhere. (I can point out the mistake, but this is a good exercise too.) In general terms, the reason is that what you did is in fact factoring the problem with 3 weighings into two problems: the first with two weighing, the second with one.

Have a look at the solution by Brian Bunday.