Sum of Prime Divisors of Polygon Vertices Triangle Count

To find the sum of all distinct prime divisors of $n$, we first determine the value of $n$ using the properties of combinations.

The total number of triangles formed by joining the vertices of an $n$-sided regular polygon is given by selecting any 3 vertices out of the $n$ available vertices: $P_n = \binom{n}{3}$

Similarly, for an $(n+1)$-sided regular polygon, the number of triangles is: $P_{n+1} = \binom{n+1}{3}$

We are given the relation: $P_{n+1} - P_n = 66$

Substituting the expressions for $P_{n+1}$ and $P_n$: $\binom{n+1}{3} - \binom{n}{3} = 66$

Using Pascal's identity, $\binom{n+1}{k} - \binom{n}{k} = \binom{n}{k-1}$, we simplify the left side: $\binom{n}{2} = 66$

Expanding the combination: $\frac{n(n-1)}{2} = 66$ $n(n-1) = 132$

Solving the quadratic equation: $n^2 - n - 132 = 0$ $(n - 12)(n + 11) = 0$

Since $n$ represents the number of sides of a polygon, it must be a positive integer, so: $n = 12$

Now, we perform the prime factorization of $n = 12$: $12 = 2^2 \times 3$

The distinct prime divisors of $12$ are $2$ and $3$.

The sum of all distinct prime divisors of $n$ is: $2 + 3 = 5$

Hence, the correct option is C.