Apologies for not being able to send the answer last weekend.
I got many correct answers for this one, and people sent different correct answers. Ones who sent correct answers include Pratik Poddar, Akash Tayal, Sandeep Singhal, Parag Dhanuka, Prakhar Prakash, Atul Tewari, Divya Mittal, Suman Saraf and Rahul Rane.
I am reproducing the answer from Sandeep, which I thought was the simplest (also sent by many others):
Weigh 50 coins against 50
If equal left out coin is odd and can be weighed against any coin.
If unequal, take any heap and 25 against 25.
If equal the other heap has the odd coin and if the other heap was heavier in the fist weighing, odd one is heavier and vice versa.
If unequal, odd one is in this heap and the odd one is heavier if this heap was heavier on the first weighing and vice versa…
Many people sent variations of this method, e.g. here is the answer from Pratik Poddar:
Divide the set of 101 coins in 3 sets:
Set A of 33 coins, Set B of 33 coins and Set C of the remaining 35 coins
Weight Set A and Set B. Without loss of generality, two cases are possible:
a) Set A == Set B
b) Set A > Set B
In the first case, we know that the counterfeit coin is in Set C, and Set A and B are all genuine coins. Using the coins of A and B, we create a 35 coin set which we know is genuine. Just weighing this against C tells us whether the counterfeit coin is heavier or lighter.
In the second case, we know that Set C is all genuine coins. Remove 2 coins from Set C. And weigh Set C against Set A, if Set A == Set C, Set B has counterfeit coin and its lighter. If Set A > Set C, Set A has counterfeit coin and is heavier.
In general, as long as you weight 26 or more coins on one side in the first weighing or (ceiling of x/2 for x coins), one can get to the answer.
Hope you enjoyed the puzzle!