longest way to rearrange students before returning to original arrangement? This is Q24 from the 2012 Intermediate 
Australian Mathematics Competition:
"A teacher has a class of twelve students. She thinks it would be a nice idea if they change desks every day, so she has painted arrows on the floor from desk to desk. Each desk has one arrow going to it and another arrow going from it. Each morning, the students pick up their books and move to the desk indicated by the arrows. By choosing her arrows carefully, the teacher has arranged it so that the longest possible time will pass before all students are back in their original desks at the same time. How many days is that?
(A) 30 (B) 35 (C) 42 (D) 60 (E) 72"
Is it right if I think that the longest arrangement is such that the student moves to the table next to him/her and does the reverse, so it would take a total of 24 days to get back the same spot?
 A: This may be over your head if you're in high school, but you want the maximum order of an element in the symmetric group on 12 letters. (In other words, you want Landau's function evaluated at 12.) This is 60, generated by the product of a 5-cycle, 4-cycle, and 3-cycle. (This is surely the way the people who wrote the test thought about this problem!)
If this is over your head, here is a way to see that 60 days is possible. Pick a group of five students and a group of four students (this leaves three remaining). Each day permute within each group cyclically, moving the students left to right, say. Since the least common multiple of 3, 4, and 5 is 60, it will take 60 days to return the students to their original arrangement.
It is more tedious to see that we can't do better than 60. One approach is brute force. We need to consider a positive integer partition of 12 whose least common multiple is as large as possible. (Note that 5+4+3=12, so the numbers 5, 4, and 3, which we used above, do indeed partition 12.) For example, another partition of 12 is 8, 3, 1, and this has an LCM of 24.
But your exam was multiple choice, and the only answer greater than 60 was 72. To see that 72 is not possible, note that an LCM of 72 would require that 8 be one number in the partition of 12. This doesn't leave enough room to get a factor of 9.
