11.3. Given $n$ coins of pairwise distinct masses and $n$ balance scales, $n>2$. In each weighing, it is allowed to choose some one scale, place one coin on each of its pans, look at the scale's reading, and then remove the coins. Some one of the scales (it is unknown which) is broken and may randomly give either a correct or an incorrect result. What is the minimum number of weighings required to definitely find the heaviest coin?
Answer. In $2 n-1$ weighings.

Solution. First, we prove that the heaviest coin can be found in $2 n-1$ weighings. More precisely, we will prove by induction on $n$ that the heaviest of $n \geqslant 2$ given coins can be determined in $2 n-1$ weighings, using three scales, one of which may be faulty.

If $n=2$, we weigh the two given coins in turn on three different scales. If the scales are in balance during one of the weighings, then these scales are faulty, which means we can determine the heavier coin by the readings of any of the other scales. If there was no balance at all, then one of the coins will outweigh at least twice - it is the heavier one, since incorrect results can only be given by one set of scales. This provides the base of the induction.

Now let $n \geqslant 3$. We choose two coins and two scales and compare these coins with each other on the first and second scales in the first two weighings. There are two possible cases: