13.20  There are 11 bags of coins and a balance scale with two pans, the scale has a pointer that can indicate which side is heavier and by how much. It is known that 1 bag contains counterfeit coins while the rest of the bags contain genuine coins. All genuine coins weigh the same, all counterfeit coins weigh the same but differ from the genuine coins. How many times at minimum must you weigh to determine which bag contains the counterfeit coins?
[Solution 1] Place 1 coin from each of the first 10 bags on the left pan of the balance, and 10 coins from the last bag on the right pan for the first weighing. Clearly, they will not balance; assume the left is lighter and the right is heavier, and record the weight difference as $a$.

Then, take $1, 2, \cdots, 10$ coins from the first 10 bags and place them on the left pan, and take 55 coins from the last bag and place them on the right pan for the second weighing. Of course, it will still be left light and right heavy, and record the weight difference as $b$. If $a: b=2: 11$, then the last bag is counterfeit; if $a: b=1: j(1 \leqslant j \leqslant 10)$, then the $j$-th bag is counterfeit. It is evident that the counterfeit can be identified with at least two weighings.
[Solution 2] Take $1, 2, 3, 4, 5$ coins from the first 5 bags and place them on the left pan of the balance, and take $1, 2, 3, 4, 5$ coins from the next 5 bags and place them on the right pan for the first weighing. If they balance, then the last bag is counterfeit; if they do not balance, then the last bag is genuine.

In the case of imbalance, assume the left is lighter and the right is heavier, and record the weight difference as $a$. Then, take 1 coin from each of the first 10 bags and place them on the left pan, and take 10 genuine coins from the last bag and place them on the right pan for the second weighing. If the left side is heavier, then the counterfeit is heavier, and the counterfeit is in one of the bags numbered $6, 7, 8, 9, 10$. If the left side is lighter, then the counterfeit is lighter, and the counterfeit is in one of the first 5 bags. Record the weight difference as $b$. When $a: b= j: 1(1 \leqslant j \leqslant 5)$, and considering whether the counterfeit is light or heavy, the bag containing the counterfeit can be determined. It is evident that the counterfeit can be identified with at least two weighings.

On the other hand, if the balance is not equal in one weighing, it is clearly impossible to determine the counterfeit. Therefore, to determine the counterfeit, at least two weighings are required.