13. (10 points) A class of 30 students goes to the movies, and their student IDs are $1, 2, \cdots, 30$; the movie tickets they hold are exactly the 1st, 2nd, ..., 30th seats in a row. Now, the movie tickets are to be distributed to these students according to the following requirements: for any two students, A and B, if A's student ID can be divided by B's student ID, then A's movie ticket number can also be divided by B's movie ticket number. How many different ways are there to distribute the movie tickets? $\qquad$
【Answer】Solution: Student No. 1 has 29 people as its multiple, so Student No. 1 can only take Movie Ticket No. 1;
Student No. 2 has 14 people as its multiple, so Student No. 2 can only take Movie Ticket No. 2;
Student No. 3 has 9 people as its multiple, so Student No. 3 can only take Movie Ticket No. 3;
Student No. 4 has 6 people as its multiple, so Student No. 4 can only take Movie Ticket No. 4;
Student No. 5 has 5 people as its multiple, so Student No. 5 can only take Movie Ticket No. 5;
Student No. 6 has 4 people as its multiple, so Student No. 6 can only take Movie Ticket No. 6;
Student No. 7 has 3 people as its multiple, so Student No. 7 can only take Movie Ticket No. 7;
Student No. 8 must be a multiple of Student No. 2 (2) and also a multiple of Student No. 4 (4), and there are 2 people as its multiple, so Student No. 8 can only take Movie Ticket No. 8;
Student No. 9 must be a multiple of Student No. 3 (3) and cannot be 6, and there are 2 people as its multiple, so Student No. 9 can only take Movie Ticket No. 9;
Student No. 10 must be a multiple of Student No. 2 (2) and also a multiple of Student No. 5 (5), and there are 2 people as its multiple, so Student No. 10 can only take Movie Ticket No. 10;
Student No. 12 must be a multiple of Student No. 3 (3) and also a multiple of Student No. 4 (4), and there is 1 person as its multiple, so Student No. 12 can only take Movie Ticket No. 12, and at the same time, Student No. 24 can only take Movie Ticket No. 24;
Student No. 14 must be a multiple of Student No. 2 (2) and also a multiple of Student No. 7 (7), and there is 1 person as its multiple, so Student No. 14 can only take Movie Ticket No. 14, and at the same time, Student No. 28 can only take Movie Ticket No. 28;
Student No. 15 must be a multiple of Student No. 3 (3) and also a multiple of Student No. 5 (5), and there is 1 person as its multiple, so Student No. 15 can only take Movie Ticket No. 15, and at the same time, Student No. 30 can only take Movie Ticket No. 30;
For the numbers after that, \([2,9]=18, 18\) must take 18, and at the same time, 27, which is a multiple of 9, can only take 27; \([4,5]=20, 20\) must take 20; \([3,7]=21, 21\) must take 21; \([3,8]=24, 24\) must take 24, and at the same time, 16, which is a multiple of 8, can only take 16; \([4,7]=28, 28\) must take 28; \([5,6]=30, 30\) must take 30, and at the same time, 25, which is a multiple of 5, can only take 25.
The numbers that have not been determined yet are 11, 22, 13, 26, 17, 19, 23, 29. 11 and 22 form a pair of multiples; 13 and 26 form a pair of multiples, and there are two ways to take them: 11 takes 11, 22 takes 22, 13 takes 13, 26 takes 26 or 11 takes 13, 22 takes 26, 13 takes 11, 26 takes 22. 17, 19, 23, 29 are large prime numbers, with no restrictions, and can be taken in any way, with \(\mathbb{A}_{4}^{4}=24\) ways, so there are \(2 \times 24=48\) ways in total.