Good $n$ -gons

Not only will we find out what $N_{20}+N_{15}$ is, but we will also find out what $N_{x}$ is for any positive integer $x$ , where $x \ge 3$ . Let's think about how we will construct such a polygon. First, we have $2015$ points to start out the first vertex of the polygon. Next, we have $x-1$ vertices left to pick. But there is a clever 1-1 correspondence between this problem and stars and bars/sticks and stones. We have a total of $2015$ vertices in a $2015$ -gon. In order to have a good $x$ -gon in which no two vertices are consecutively used from the $2015$ -gon, we can simplify this down into $a_{1}+a_{2}+...+a_{x}=2015$ where all the addends are greater than or equal to 2. $a_{1}, a_{2}, ... a_{x}$ all represent the number of vertices from the $2015$ -gon that are between each side of the $x$ -gon. Now, we can simplify down our equation into $a_{1}+a_{2}+...+a_{x}=2015-x$ where $a_{1}, a_{2}, ... , a_{n}$ are all positive integers. Using stars and bars/sticks and stones, we have $2015-x-1=2014-x$ stars/stones and $x-1$ bars/sticks. Therefore, we have $\dbinom{2014-x}{x-1}$ such good $x$ -gons. But all good $x$ -gons are overcounted. Consider such a good $x$ -gon. We had $2015$ points to start the first vertex off at. So therefore, we could have started our good $x$ -gon off at any of the vertices of our $x$ -gon. We need to divide our final answer by $x$ to account for the rotations. So our final answer is $\frac{\dbinom{2014-x}{x-1}}{x}$ . Plugging in for $x=20$ and $x=15$ , we have $\frac{2015 * \dbinom{1999}{14}}{15}+\frac{2015 * \dbinom{1994}{19}}{20}$ . In Note 1 as stated in the problem, you need a CAS to find the digit sum of this, but I do not know what CAS stands for and instead went to Wolfram Alpha to find out the digit sum. I got 194.