5. On each of five cards, a number is written. The cards lie on the table numbers down. We can, by paying a ruble, point to any three cards, and we will be told the sum of the numbers written on them. For what minimum price can we surely find out the sum of all five numbers? (I. Rubanov)
Answer: For 4 rubles. Solution: Let the numbers written be $a, b, c, d, e$. We ask about the sums $a+b+c, a+b+d$, $a+b+e, c+d+e$. Then, by adding the first three sums and subtracting the fourth from the result, we get $3(a+b)$, then $a+b$ and, by adding $c+d+e$ to the result, the sum of all five numbers.

Suppose we managed to get by with three questions. Let's call the presence of a card in a question an occurrence. If there is a card with three occurrences, then we increase the number on it by 2, and decrease all other numbers by 1. Then the answers to all three questions will not change, but the sum of all numbers will decrease by 2. Therefore, it is impossible to determine this sum with such three questions. This means that each card has no more than two occurrences, and this is only possible if four cards have two occurrences each, and one has one. By changing the notation if necessary, we can assume that the single occurrence is on card $e$, and one of the questions is $-a+b+c$. Then in another question without card $e$, two of the three cards $a, b, c$ must be present. By changing the notation if necessary, we can assume that this is the question $a+b+d$. Then in the third question, neither of the cards $a$ and $b$ can be present, so this is the question $c+d+e$. But then, by increasing each of the numbers $a$ and $e$ by 2 and simultaneously decreasing each of the numbers $b, c$, and $d$ by 1, the answers to all three questions will not change, but the sum of all five numbers will increase by 1. Therefore, in this case, it is also impossible to determine the sum of all five numbers.