, 2016

Solution:

In order for $c_{n} \neq 0$, we must have $\operatorname{gcd}(n, 210)=1$, so we need only consider such $n$. The number $n^{c_{n}}-1$ is divisible by 210 iff it is divisible by each of $2,3,5$, and $7$, and we can consider the order of $n$ modulo each modulus separately; $c_{n}$ will simply be the LCM of these orders. We can ignore the modulus 2 because order is always 1. For the other moduli, the sets of orders are
$$\begin{array}{rr} a \in\{1,2\} & \bmod 3 \\ b \in\{1,2,4,4\} & \bmod 5 \\ c \in\{1,2,3,3,6,6\} & \bmod 7 \end{array}$$
By the Chinese Remainder Theorem, each triplet of choices from these three multisets occurs for exactly one $n$ in the range $\{1,2, \ldots, 210\}$, so the answer we seek is the sum of $\operatorname{lcm}(a, b, c)$ over $a, b, c$ in the Cartesian product of these multisets. For $a=1$ this table of LCMs is as follows:

which has a sum of $21+56+28+56=161$. The table for $a=2$ is identical except for the top row, where $1,3,3$ are replaced by $2,6,6$, and thus has a total sum of 7 more, or 168. So our answer is $161+168=\mathbf{329}$.

This can also be computed by counting how many times each LCM occurs:
- 12 appears 16 times when $b=4$ and $c \in\{3,6\}$, for a contribution of $12 \times 16=192$;
- 6 appears 14 times, 8 times when $c=6$ and $b \leq 2$ and 6 times when $c=3$ and $(a, b) \in$ $\{(1,2),(2,1),(2,2)\}$, for a contribution of $6 \times 14=84$;
- 4 appears 8 times when $b=4$ and $a, c \in\{1,2\}$, for a contribution of $4 \times 8=32$;
- 3 appears 2 times when $c=3$ and $a=b=1$, for a contribution of $3 \times 2=6$;
- 2 appears 7 times when $a, b, c \in\{1,2\}$ and $(a, b, c) \neq(1,1,1)$, for a contribution of $2 \times 7=14$;
- 1 appears 1 time when $a=b=c=1$, for a contribution of $1 \times 1=1$.
The result is again $192+84+32+6+14+1=329$.