Task 11.2. On the table, there are 30 coins: 23 ten-ruble coins and 7 five-ruble coins, with 20 of these coins lying heads up and the remaining 10 tails up. What is the smallest $k$ such that among any $k$ randomly selected coins, there will definitely be a ten-ruble coin lying heads up?
Answer: 18.

Solution. If you choose 18 coins, then among them there will be no more than 10 lying heads down, so at least 8 coins will be lying heads up. Among these coins, no more than 7 will be five-ruble coins, so at least one will be a ten-ruble coin, which is what we need.

On the other hand, if initially on the table there are 7 five-ruble coins heads up, 10 ten-ruble coins heads down, and 13 ten-ruble coins heads up, then among 17 coins there might be only coins of the first two types, so 17 coins (or fewer) might not be enough.