2017 students in a circle

If there are $3^n$ students sitting around the circle, then after the first round of counting there remain $3^{n-1}$ students and student $S_0$ will count the same number $1$ in the second round.

So after $n$ rounds, $S_0$ will be the last student remaining.

Now suppose that there are $2017$ students.

Since $3^6 = 729 < 2017 < 3^7 = 2187$ , we can reduce our problem to the previous easy case when there are $3^6$ students left; then student $S_k$ will be the student who will count $1$ first among these $3^6$ students.

We need to remove $2017 - 729 = 1288 = 2\times 644$ students.

These correspond to $644$ groups of three students (with two left out from each group).

So we need $644\times 3 = 1932$ students sitting before $S_k$ , and hence $k = \boxed{1932}$ .