10.9. A magician and an assistant are going to perform the following trick. They have $n \geqslant 3$ cards numbered $1,2, \ldots, n$, and a row of $n$ cells the size of a card. The back sides of all cards are indistinguishable. The audience places cards 1 and 2 on any two spots; the assistant, seeing this, places the remaining cards on the free spots. Then all cards are flipped numbers down, and the magician enters. He flips one of the cards, and then the audience flips another. After this, the magician must correctly identify card 1 and correctly identify card 2. For which $n$ can the magician and the assistant agree in advance so that the trick is guaranteed to succeed

(I. Bogdanov, K. Knop)
Answer. For all $n$.

Solution. Let's present one of the possible agreements between the magician and the assistant. Mark the leftmost position with a * (the magician will flip the card at this position). Number all other positions from 1 to $n-1$; the magician and the assistant will consider these positions to be arranged in a cycle, meaning position 1 follows position $n-1$.

If the card at position * is card 1 or 2 (let's assume it is card 1 for definiteness), the assistant lays out all their cards in ascending order in the cycle after card 2. Then the magician, seeing the card at position *, can restore the cyclic order of all other cards (i.e., their order up to a cyclic shift), and from any card opened by the audience, the magician can determine the positions of all cards.

Now suppose cards 1 and 2 are in the cycle. The assistant counts the number of positions $i$ that need to be moved in the cycle to get from card 1 to card 2; then $1 \leqslant i \leqslant n-2$. The assistant then places card $i+2$ at position *. The other cards are laid out in ascending order in the cycle after card 1, skipping the position of card 2. Seeing the card at position *, the magician learns the number $i$. Using this, he can again restore the cyclic order of the cards in the other positions, and from any card opened by the audience, he can determine the positions of all other cards.

Remark. There are other strategies. However, all working strategies must have the following properties.

As in the solution above, let * denote the position where the magician will flip the card. There are $n(n-1)$ possible placements of cards 1 and 2; thus, there are as many possible placements of all cards after the assistant's actions. For any $1 \leqslant i \leqslant n$, card $i$ must end up at position * in exactly $n-1$ of these placements. Moreover, in these $n-1$ placements, any card $j \neq i$ must be in $n-1$ different positions (other than *).

(The easiest way to achieve the latter property is to make these placements differ by cyclic shifts of the cards in positions other than *, as in the solution above; however, this can be achieved by other methods as well.)

Comment. No points are awarded for an answer without justification and/or for considering several particular cases (for small values of $n$).

If the solution only outlines the structure of the magician's and assistant's strategy described in the remark above (with or without a proof that it must be exactly this way) -2 points.

If the solution provides a strategy that works for an infinite set of values of $n$, with a proof that it works, at least 3 points. If the strategy does not work for an infinite set of values of $n$ - exactly 3 points.