I suppose this can be done in a simpler way than discussed above!

Make 2 stacks each time and put them on the balancing pan. Lets name the two sides of balancing pan as A&B. This is how it can be done-

1. Make 2 stacks of 13 coins each and put on A&B.

Now if U'r lucky, both the pans would weigh same & thus the leftover coin is ur choice!

BUT, If Pan B is heavy-

2. Then take the 13 coins from pan B, divide in 6 coins each & place on A&B again.

IF u'r lucky then A&B would weigh equal & the left over coin is heavier one!

BUT If either side of pan is heavy (lets take A this time!)

3. Take the 6 coins from A, divide into 3 coins each & place on A&B again!

No luck here buddy

! One of either side of pan would be heavy

(lets take B this time!)

4. Take the 3 coins, keep one coin on each side of A&B.

WOW....here u get ur coin...If one side gets heavy, then the heavier side is ur coin!...& if both weigh equal, then the leftover coin is the heavy one!!!

I suppose this is a much simpler solution than the ones dealing in making 3 stacks

& comparing them again & again <_< . Those solutions really would be quite painstaking, confusing & require lot of Permutations & Combinations.