14. 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 no more than 11 weighings on a balance scale without weights?

Note. The problem assumes that all cubes can be aluminum, but they cannot all be duralumin (since if all the cubes were of the same weight, we would have no way to determine whether they are aluminum or duralumin without this condition).
14. Let's place one cube on each pan of the scales (first weighing). In this case, two different scenarios may occur.

$1^{\circ}$. During the first weighing, one pan of the scales tips. In this case, one of the two weighed cubes is definitely aluminum, and the other is duralumin. Next, we place these two cubes on one pan of the scales, and on the other pan, we sequentially place pairs of the remaining cubes (we randomly divide the 18 remaining cubes into 9 pairs). If any pair of cubes tips the scales, it means that both cubes in the second pair are duralumin; if the first pair tips the scales, it means that both cubes in the second pair are aluminum; if both pairs have the same weight, it means that the second pair also contains one aluminum and one duralumin cube. Thus, in case $1^{\circ}$, we can determine the number of duralumin cubes using 10 weighings (one weighing and 9 more).

$2^{\circ}$. During the first weighing, the scales remain in balance. In this case, the cubes of the first pair are either both aluminum or both duralumin. Next, we place these two cubes on one pan of the scales, and on the other pan, we sequentially place pairs of cubes from the remaining 18. Suppose the first $k$ of these pairs have the same weight as the initial pair, while the $(k+1)$-th pair has a different weight. (If $k=9$, all the cubes have the same weight, and thus there are no duralumin cubes at all; the case $k=0$ is no different from the general case). For definiteness, let's assume that the $(k+1)$-th pair is heavier than the initial pair (the reasoning would be similar if the $(k+1)$-th pair were lighter). In this case, the first two cubes, and therefore the cubes of the $k$ pairs that have the same weight as them, are definitely aluminum. Thus, we have performed $1+(k+1)=k+2$ weighings and identified $k+1$ pairs of aluminum cubes. Now, we place one cube from the last weighed pair on each pan of the scales (the $(k+3)$-th weighing). If both cubes have the same weight, they must both be duralumin; otherwise, one is aluminum and the other is duralumin. In both cases, after $k+3$ weighings, we can identify a pair of two cubes, one of which is aluminum and the other is duralumin. Using this pair, we can determine the number of duralumin cubes among the remaining $20-2(k+2)=16-2k$ cubes with $8-k$ weighings, similar to how we proceeded in case $1^{\circ}$. The total number of weighings in case $2^{\circ}$ will be $k+3+(8-k)=11$.