5.66 A deck of cards has a total of 54 cards. The magician divides them into several piles, and the audience writes a natural number on each card, which is equal to the number of cards in the pile where the card is located. Then the magician mixes the cards in a special way and divides them into several piles again. The audience writes a natural number on each card again, which is also equal to the number of cards in the pile where the card is now located. This process continues. How many times does this process of dividing piles and writing numbers need to be repeated at least, so that the (unordered) number arrays written on each card are all different?
[Solution] From the right figure of the previous problem, it can be seen that after two rounds of dividing the cards and writing numbers, the two numbers written on the cards are the same (disregarding the order) for at most two cards, with the corresponding cells one above the main diagonal and the other below the main diagonal and symmetric about the main diagonal. During the 3rd division, the 30 numbers on and above the main diagonal can be divided into one pile, and the 24 numbers below the main diagonal can be divided into another pile. Then, the cards in the two piles can be written with the 3rd number 30 and 24, respectively. In fact, the 3rd division will not eliminate the differences of the original number pairs, and it will distinguish the two cards originally as $(m, n)$ and $(n, m)$, turning them into $(m, n, 30)$ and $(n, m, 24)$, where $1 \leqslant m \leqslant n \leqslant 9$. It can be seen that appropriate 3 divisions and writing numbers can meet the requirements.

Next, we prove that two divisions and writing numbers are insufficient. If not, suppose that two divisions and writing numbers can meet the requirements of the problem, where the first division results in a pile with the most cards, say $k$ cards, and let this be the 1st pile. At this point, each natural number written on a card is one of $1,2, \cdots, k$. During the 2nd division, suppose the pile with the most cards has $m$ cards. If $m > k$, then among the $m$ cards, at least two cards have the same first number, thus the two numbers are the same, which does not meet the requirements; if $m < k$, a similar discussion leads to a contradiction, so it must be that $m = k$. Therefore, the two divisions and writing numbers can be represented by a $k \times k$ grid table as in the previous proof. Since the 1st pile in the 1st division has $k$ cards, the $k$-th row of the table has $k$ "×"s. Similarly, the $k$-th column of the table also has $k$ "×"s. They are in positions symmetric about the main diagonal. The numbers on the cards corresponding to the cells in symmetric positions are $(n, k)$ and $(k, n)$, which do not meet the requirements of the problem.