6.1. Coins come in denominations of 50 kopecks, 1 ruble, 2 rubles, 5 rubles, 10 rubles. $V$ has several coins in his wallet. It is known that no matter which 20 coins are taken out of the wallet, there will be at least one 1-ruble coin, at least one 2-ruble coin, and at least one 5-ruble coin. What is the maximum number of coins in the wallet for this to be possible?
Answer: 28

Solution. Example: 9 coins of 1 ruble, 9 coins of 2 rubles, 9 coins of 5 rubles, and 1 coin of 10 rubles. Note that the wallet contains a total of $9+9+1=19$ coins that are not 1 ruble, so among any 20 coins, there will definitely be a 1-ruble coin. Similarly, this can be verified for all other denominations.

Estimate: Suppose there are $x$ coins in the wallet. Since the number of coins not worth 1 ruble is no more than 19 (otherwise, there would be 20 coins without a 1-ruble coin), the number of 1-ruble coins must be at least $x-19$. Similarly for 2-ruble and 5-ruble coins. Therefore, the total number of coins is at least $3(x-19)$. We get the inequality $x \geqslant 3 x-57$, from which $57 \geqslant 2 x, 28.5 \geqslant x$. Since $x$ is an integer, it does not exceed 28.