Four friends Adam, Mojmír, and twins Petr and Pavel earned a total of 52 smileys in math class, each getting at least 1. The twins together have 33, but the most successful was Mojmír. How many did Adam get?

(M. Volfová)
All smileys amount to 52, while the twins got 33 and Adam at least one. For Mojmír, there are at most $52-33-1=18$ smileys left. For him to have the most of all, each of the twins can have at most 17 smileys. This means that Petr got exactly 17 and Pavel 16, or vice versa. If one had less than 16, the other would have to have more than 17 to make a total of 33. From this, it also follows that Mojmír could not have gotten less than 18 smileys to have more than each of the twins. Therefore, Mojmír got exactly 18 smileys, and one smiley is left for Adam :-).

Another solution. We know that the twins got a total of 33 smileys, and each got at least one. If Petr got 32 smileys and Pavel one, Mojmír would have to get at least 33 to have the most. Then, all together with Adam, they would have at least $33+33+1=67$ smileys, which is not possible because from the problem statement we know they have 52 in total. Similarly, if Petr got 31 smileys and Pavel 2, Mojmír would have to have at least 32, together with Adam then $33+32+1=66$, which is still too much. ... By the same reasoning, all possibilities for the distribution of smileys between the twins can be excluded except for the following case: Petr 17, Pavel 16 (or vice versa), then Mojmír 18 and Adam 1.