For positive integers $n$, let $c_n$ be the smallest positive integer for which $n^{c_n} -1$ is divis by $210$. What is $c_1 + c_2 + \dots + c_{210}$? 
For positive integers $n$, let $c_n$ be the smallest positive integer for which $n^{c_n} -1$ is divisible by $210$, if such a positive integer exists, and $c_n = 0$ otherwise. What is $c_1 + c_2 + \dots + c_{210}$?

$n^{c_n}-1$ being divisible by $210$ is the same as saying that $n^{c_n} \equiv 1 \pmod {210}$ and since $c_n$ is the smallest such integer we have that $\operatorname{ord}_{210}(n)=c_n$.
We also have that $\operatorname{ord}_{210}(n) \mid \varphi(210)$ that is $c_n \mid 48$.
So $c_n$ can only be something of $1,2,3,4,6,8,12,16,24,48$.
How can I get further with the problem? I'm a bit stuck here.
 A: Hmmm. OP never responded to further prompting, but I feel like the problem is interesting enough deserve a solution.
As above, we have $c_n = \text{ord}_{210}(n)$. (I'm just going to use $o_m(n)$ for the order for simplicity from here forward.) We know only $\varphi(210) = 48$ values of $n$ will in fact have a multiplicative order, and that for those that have one,
$$o_{210}(n) = \text{lcm}[o_2(n),o_3(n),o_5(n),o_7(n)]$$
The problem definition tells us that for $n$ that aren't units of $\mathbb{Z}/210\mathbb{Z},c_n=0$. Then:

*

*We can ignore modulo $2$, since the order will always be either $1$ or undefined.

*In modulo $3$, $o_3(n) = n \bmod 3$.

*In modulo $5$, $o_5(n) = 1,4,4,2$ for $n \equiv 1,2,3,4 \pmod 5$ respectively.

*In modulo $7$, the orders are $1,3,6,3,6,2$ for the consecutive residues.

Sadly, there's no elementary way to sum that up. So instead we have:
$$\sum^{210}_{n=1} c_n = \sum_{1 \le n \le 210}^{(n,210) = 1} \text{lcm}[n \bmod 3, o_5(n), o_7(n)]$$
Of course, this is best found via a computer program, but it you have to to it by hand, at least everything is in nice patterns.
A: Building on Eric Snyder's answer and comments, it's easier here to use the prime power structure of the multiplicative group $(\mathbb{Z}/210\mathbb{Z})^\times$:
$$ (\mathbb{Z}/3\mathbb{Z})^\times \times (\mathbb{Z}/5\mathbb{Z})^\times \times (\mathbb{Z}/7\mathbb{Z})^\times \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z} \cong (\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2^2\mathbb{Z})\times (\mathbb{Z}/3\mathbb{Z}).$$
As Mastrem pointed out initially, we just want to sum the orders of the elements of this group.  We factor any such order into the power of $2$ arising from the $2$-group factor and the power of $3$ arising from the $3$-group, and the sums of those orders factor through much like summing divisors.  More generally, if $\sigma(G)$ is the sum of the orders of $G$, and $|G|,|H|$ are relatively prime, then $\sigma(G\times H) = \sigma(G) \sigma(H)$.  So we can solve each of these factors separately.
For $\mathbb{Z}/3\mathbb{Z}$, there is one element of order $1$ and two of order $3$, so $\sigma(\mathbb{Z}/3\mathbb{Z}) = 7$.
For $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2^2\mathbb{Z}$, the maximum order is $4$.  There is one element of order $1$, and the only way to get order $4$ is to take one of the two generators of $\mathbb{Z}/2^2\mathbb{Z}$, with the other components being arbitrary.  That makes $8$ elements of order $4$ and the other $7$ have order $2$.  So $\sigma(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2^2\mathbb{Z}) = 4\cdot 8 + 2\cdot 7 + 1 = 47$.
This yields the expected answer of $\sigma((\mathbb{Z}/210\mathbb{Z})^\times) = 7\cdot 47 = 329$.
