Puzzle - reordering people in a line Eighty players numbered I through $80$ are standing in a row, one behind the other, in the
increasing order of their numbers. The physical director of the players performed eight successive
inspections of the players and in each of the respective inspections he sent the first $10$, $20$, $30$, $40$,
$50$, $60$, $70$ and $80$ players, from the front of the row to the end of the row. Each time the players
being sent back one after another. After these eight rounds of inspections what is the position of
the player numbered $5$?

Every time I solve it I gets confused, but still is there any general solution to this problem
If I understand this correctly...

Sending one player to the back moves player $i$ to position $i - 1$ (wrapping around by adding $80$ as needed).  The director does it $10 + ... + 80 = 360$ times, so this sends $5$ to position $5 - 360$, which wraps around to $5 - 360 + 5 * 80 = 5 + 40 = 45$.
from logical point of view as i am seeing it and if i get your question right , it will be like

Original Position : 10 20 30 40 50 60 70 80

first iteration : 80 70 60 50 40 30 20 10

2nd iteration : 10 20 30 40 50 60 70 80

3rd iteration : 80 70 60 50 40 30 20 10

4th iteration : 10 20 30 40 50 60 70 80

5th iteration : 80 70 60 50 40 30 20 10

6th iteration : 10 20 30 40 50 60 70 80

7th iteration : 80 70 60 50 40 30 20 10

8th iteration : 10 20 30 40 50 60 70 80

so the last one is the same , the player position will be same.comment if this is correct