Problem 7. The magician has a deck of 52 cards for the game. The audience wants to learn the order of the cards in the deck (regardless of whether it is face up or face down). They can ask how many cards there are between two specific cards (for example, between the ace of spades and the nine of hearts). One of the audience members knows the order of the cards. What is the minimum number of questions he needs to ask the magician so that the other audience members can understand the order of the cards?
Solution: We will prove that the number of questions is at least 34.

The first audience member asks a question about the two end cards. The magician's answer is 50 and shows everyone that these cards are the end ones. Let's call one the first and the other the 52nd. We need to determine the numbers of all the other cards. Let's call the second card a hole and with the second question ask about the two adjacent cards to the hole, i.e., the first and third cards. The answer is 1 and uniquely determines the position of the third card. Then we continue to ask questions in pairs:

- for odd-numbered questions, we ask about the two end cards that have not yet been mentioned (one of them is a hole, the other is not);
- we determine a new hole, which is the unmentioned card adjacent to the non-hole;
- for the next (even) question, we ask about the two cards adjacent to the hole.

We notice that the hole first appears closer to the beginning, then closer to the end, and so on. In the first pair of questions, cards 1, 52, and 3 are mentioned; in the second pair, cards 2, 51, and 49; in the third pair, cards 4, 50, and 6, and so on. After the answers to the next pair of questions, there are two possible arrangements for the three mentioned cards: the main (the actual one) and the side (where the end cards swap places and the middle one moves accordingly). For example, after the answers to questions 3 and 4, it is clear that the second triplet is 2, 51, and 49 or 2, 51, and 4.

Main variant $\quad a b a \quad \ldots \quad b \smile b a$

Side variant $a b a b$... $b a$

(Cards from one triplet are marked the same.)

This uncertainty will disappear after the answer to the next (fifth) question (about cards 4 and 50). This is because in the side variant, the number of cards between the previously unmentioned end cards is at most 44 and is less than the number in the main one (which is 45). Therefore, by asking 33 questions, we will clarify the position of 48 cards. The last, 34th question, we will ask about the 25th and 26th cards:

abacdcefeghgijiklkmnmoPoQQ $P Q$ Pnonlmljkjhihfg dedbcba

(The penultimate and last triplet are marked with $P$ and $Q$, respectively.)

From this question, the position of the last triplet and the only remaining card is uniquely determined.

To prove that fewer questions are not sufficient, we will divide all the cards into 52 groups of one card each. When a question is asked about two cards from different groups, we combine the two groups into one. It is clear that each question reduces the number of groups by at most 1. If no more than 33 questions are asked, there will be no fewer than $52-33=19$ groups. Groups with 3 cards are no more than 17 (since $18 \cdot 3 = 54 > 52$). Therefore, either there will be two groups with one card each, or a group with exactly two cards. In both cases, by swapping the positions of these two cards, the answers will not change; this means that the sequence of cards cannot be restored.