Problem: Let $R$ be a unit square region and $n \geq 4$ an integer. A point $X$ in the interior of $R$ is called n-ray partitional if there are $n$ rays emanating from $X$ that divide $R$ into $n$ triangles of equal area. How many points are $100$ -ray partitional but not $60$ -ray partitional?
$\textbf{(A)}\ 1500 \qquad \textbf{(B)}\ 1560 \qquad \textbf{(C)}\ 2320 \qquad \textbf{(D)}\ 2480 \qquad \textbf{(E)}\ 2500$

Solution: For the sake of simplicity, let $R$ be a $60 \times 60$ square and set the bottom-left point as the origin. Then, $R$ has vertices: \[(0,0), (60,0), (60,60), (0,60).\]
Now, let a point in the square have coordinates $(x, y).$
In order for the point to be $100-$ ray partitional, we must be able to make $100$ triangles with area $60^2/100 = 36.$ For it to not be $60$ -ray partitional, we cannot make $60$ triangles with area $60^2/60 = 60.$
When we draw such a triangle, it's base will always be on one of the sides of the square. If it is on the bottom side, then the height must be $y$ . So, the base of each triangle must be $\frac{72}{y}.$ So, there will be $\frac{60}{\frac{72}{y}} = \frac{60y}{72}$ total triangles on that side.
If the triangle is on the right side of the square, then the height has to be $60 - x.$ So, the base will be $\frac{72}{60-x}$ and there will be $\frac{60 (60-x)}{72}$ triangles.
If the triangle is on the top, the height will be $60-y$ , the base will be $\frac{72}{60-y}$ and there will be $\frac{60 (60-y)}{72}$ triangles along this side.
The triangles on the left will have height $x$ , base $\frac{72}{x}$ and $\frac{60x}{72}$ such triangles must exist.
Simplifying, we get $\frac{5y}{6}, \frac{5 (60-x)}{6}, \frac{5(60-y)}{6}, \frac{5x}{6}$ triangles on the respective sides of the square.
All of these numbers must be integers. Let $x = 60a$ and $y = 60b$ where $0 < a, b < 1.$
Then, the amounts become: \[50a, 50b,  50 - 50a, 50 - 50b.\]
As long as $50a$ and $50b$ are integers, $50 - 50a$ and $50 - 50b$ will also be integers.
For this to happen, $a$ and $b$ must be of the form $\frac{x}{50}$ where $1 \le x \le 49.$ So, we have a total of $49^2 = 2401$ points that are $100$ -ray partitional.
Now, we must calculate the number of $100$ - ray partitional points that are also $60$ -ray partitional.
Using a method similar to before, if a point is $60$ -ray partitional, then we must be able to make $30a, 30b, 30 - 30a, 30 -30b$ triangles on the different sides.
So, $30a$ and $30b$ must be integers. This means $a$ and $b$ have to be of the form $\frac{y}{30}$ where $0<y<30.$
If a point is both $100$ -ray partitional and $60$ -ray partitional, then it can be written as \[\frac{x}{50} =  \frac{y}{30}\] . Note that whenever $x$ is divisible by $5$ , a $y$ will always exist to satisfy the above equation.
So, $x$ has to be in the range $(0, 50)$ and must be divisible by $5$ . There are $9$ possibilities, namely $5, 10, 15, 20, 25, 30, 35, 40, 45.$
The $x$ -coordinate of the point and the $y$ -coordinate of the point can each use any of these $9$ possibilities, giving $9^2 = 81$ numbers that are both $100$ -ray partitional and $60$ -ray partitional.
Overall, we are left with $2401 - 81 = \boxed{2320}$ solutions.

Answer:  2320