36 ! That seems to be too much. You will never be able to solve the problem if you try a new line of thinking. This problem is just a modification of the previous problem.
Let us denote the coins by C1, C2, ..... , C36. Make 12 groups containing 3 coins each. The group can be denoted by (C1, C2, C3), (C4, C5, C6), ...... , (C34, C35, C36). Consider each group as a single coin. Following same procedure as in problem-2 find the defective coin and its defect in 3 chances. Now we have one chance left. We get the following cases.
Case-1:
We have a group of 3 coins say C1, C2 and C3 containing one heavy coin.
Put
C1 in P1 and C2 in P2. If P1 goes down then C1 is heavy. If P2 goes down
then C2 is heavy. If they balance each other then C3 is heavy.
Case-2:
We have a group of 3 coins say C1, C2 and C3 containing one light coin.
Put
C1 in P1 and C2 in P2. If P1 goes up then C1 is light. If P2 goes up then
C2 is light. If they balance each other then C3 is light.