This is problem 188.8.131.52 from 66 MOSCOW MATHEMATICAL OLYMPIADS (author: N. Konstantinov).
"A snail crawls along a straight line, always forward, at a variable speed. Several observers in succession follow its movements during $6$ minutes. Each person begins to observe before the preceding observer finishes the observation and observes the snail for exactly one minute. Each observer noticed that during his (her) minute of observation the snail has crawled exactly $1$ meter. Prove that during $6$ minutes the snail could have crawled at most $10$ meters."
I have also found a solution that reads as follows:
"We will show that if there are $n ≥ 10$ people watching, then the snail can crawl at most $n$ metres. Certainly, the snail cannot crawl further than this, since each person watches the snail crawl exactly one metre. To show that the snail can in fact crawl $n$ metres, suppose that all $n$ people watch the snail remain stationary during the first $(n − 10)/(n − 1)$ hours. After that, the n people take turn to watch the snail crawl one metre in $9/(n−1)$ hours. Then each person has watched the snail for $(n − 10)/(n − 1) + 9/(n − 1) = 1$ hour and the snail has crawled for a total of $(n − 10)/(n − 1) + 9n/(n − 1) = 10$ hours, during which it travels $n$ metres."
Can anyone help me understand the solution? First of all, the solution mentions hours, I guess the guy means "minutes", but let's disregard this. What does it have to do with people watching? What is $(n − 10)/(n − 1)$? ...and all the rest, of course!