20.16. Some of the 20 metal cubes, identical in size and appearance, are aluminum, the rest* are duralumin (heavier). How can you determine the number of duralumin cubes using 11 weighings on a balance with two pans and no weights?
20.16. Let's put one cube on each pan of the balance. There are two possible cases.

Case 1. One of the cubes turned out to be heavier in the first weighing.

In this case, one of the selected cubes is aluminum, and the other is duralumin. Place the selected cubes on one pan and compare them with the remaining cubes. Specifically, divide the remaining 18 cubes into 9 pairs and place them on the other pan one by one. Each time, we can determine how many duralumin cubes are in the pair. Indeed, if the reference pair is lighter, then we placed two duralumin cubes; if the reference pair has the same weight, then we placed one aluminum and one duralumin cube; if the reference pair is heavier, then we placed two aluminum cubes. Thus, in the first case, 10 weighings are sufficient.

Case 2. The cubes turned out to be of equal weight in the first weighing.

In this case, either both selected cubes are aluminum or both are duralumin. Place the selected cubes on one pan and sequentially compare them with the remaining cubes. Suppose the first $k$ pairs have the same weight, and the $(k+1)$-th pair has a different weight. (If $k=9$, then all cubes have the same weight, so there are no duralumin cubes.) Let's assume for definiteness that the $(k+1)$-th pair is heavier. Then the first two cubes and the cubes of the first $k$ pairs are aluminum. Place one cube from the $(k+1)$-th pair on each pan of the balance. If these cubes have the same weight, then they are both duralumin. If the cubes have different weights, then one is aluminum and the other is duralumin. In both cases, we can form a pair of cubes, one of which is aluminum and the other is duralumin. We can compare the remaining pairs of cubes with this pair, as in the first case. The total number of weighings in the second case is 11.