Van has equal-sized balls, of which 2 are red, 2 are white, and 2 are blue. Among the monochromatic balls, one is made of gold, and the other is made of silver. What is the minimum number of weighings needed to select the three gold balls using a two-pan balance? (Balls made of the same material have the same weight.)
We claim that the three gold balls - these are the heavier ones - can be identified after two weighings, but one weighing is insufficient for this. First, we will prove this latter statement.

Let the red balls be denoted by $p_{1}$ and $p_{2}$, the white ones by $f_{1}$ and $f_{2}$, and the blue ones by $k_{1}$ and $k_{2}$. The numbers will allow us to refer to one or the other of the balls of the same color.

The balls numbered 1 could have been made of gold or silver independently, which means we have $2^{3}=8$ possibilities for the materials of the individual balls. Now, the result of a weighing with a two-pan balance can be of three types, and each of the three outcomes represents a suitable subset of the eight possibilities. Since a single weighing cannot distinguish the eight possibilities, at least two weighings are needed to solve the problem.

It is also clear that two weighings, with $3 \cdot 3=9$ possible outcomes, could in principle be sufficient for the solution, but of course, it does not in itself guarantee the existence of a good procedure consisting of two weighings.

In any case, we must approach it so that the first weighing does not have an outcome that leaves us with at least four further possibilities, since these cannot be distinguished with a single further weighing.

Let's place two balls on each pan for the first weighing, specifically $p_{1}$ and $f_{1}$ on one pan, and $p_{2}$ and $k_{1}$ on the other. The table below shows what results the various possibilities lead to. In the table, $e$ and $a$ denote that the corresponding ball was silver or gold, respectively.

|  | $p_{1}$ | $f_{1}$ |  | $p_{2}$ | $k_{1}$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 1 | $e$ | $e$ | $$ | $e$ | $e$ |
| 6 | $a$ | $e$ | $=$ | $e$ | $a$ |
| 7 | $a$ | $a$ | $>$ | $e$ | $e$ |
| 8 | $a$ | $a$ | $>$ | $e$ | $a$ |

It can be seen that whatever the result of the weighing, the number of cases that still need to be distinguished is at most three, specifically two in the case of balance, and three otherwise.

If the first weighing resulted in a balance, for example, by weighing the two red balls again, if $p_{1}<p_{2}$, then we are faced with the 3rd, and in the opposite case, the 6th case - according to the numbering of the above table - and thus we know the material of each ball after the second weighing.

If there was no balance, then by symmetry, it can be assumed that the pan containing the 1st red ball was the lighter one. Then $p_{1}$ is silver, $p_{2}$ is gold, and it is not possible that $f_{1}$ is gold and $k_{1}$ is silver. The three possibilities are: $f_{1}$ is silver and $k_{1}$ is silver, $f_{1}$ is silver and $k_{1}$ is gold, and both $f_{1}$ and $k_{1}$ are gold.

Notice that the combined weight of $f_{1}$ and $k_{1}$ is different in all three cases, so by placing $f_{1}$ and $k_{1}$ on one pan and the two red balls on the other, the pan with the "mixed" balls will be lighter, the same, or heavier than the pan with the red balls, depending on whether the first, second, or third of the remaining three possibilities holds. This completes the proof.

Remark. It can be shown that a single weighing does not make it possible to select the heavier balls even in a "lucky" case, that is, no matter how we perform a single weighing, none of its outcomes provide sufficient information. It is also easy to show that, in essence, the above procedure is the only one that selects the gold balls with two weighings.