Pick's theorem application In a game let us say that we have $3$ ways to score points, getting $1,2 \text{ or } 3$ points. I make a total of $30$ points. What are the various ways I can make $30$ points? 
If we plug in values, we have $a + 2b + 3c = 30$ and we have to find ordered pairs $a,b,c$ where $a,b,c$ are whole numbers. However i saw another way of solving this question, using Pick's theorem. I found it better because plugging in would be difficult if it would have been $300$ in place of $30$.
According to the solution I read, the number of $2$s and $3$s will affect the number of $1$s, so we can make represent the $2$s on the x-axis and the $3$s on the y-axis and vice-versa. So how can I use the Pick's theorem to solve this question?

Your problem is that in Pick's Theorem the boundary points count only as 1/2 (not 1) but for you the boundary solutions are as good as the interior ones. Therefore, area of that triangle will not directly give you the number of solutions. You must count the boundary solutions separately. Namely, if area = $S$ and number of boundary solutions is $b$ then your answer $x$ would be $x = S+1+b/2$. In this specific case it is easy to find $b$ because you know that there are $N/2+N/3+N/6$ solutions on the sides of the triangle ($N=30$ in your case and let us assume for the moment that $N$ is divisible by both 2 and 3). So we get
$$
b = N/2+N/3+N/6, \quad x = S + 1 + (N/2+N/3+N/6)/2
$$
But in this case you really do not need to involve Pick's Theorem at all because you have a very simple case of rectangular triangle with its sides on the axes (a subcase of PT which is actually used to prove the theorem). Number of integer points covered by rectangle $10\times15$ is $11\times16=176$, the number of points on the diagonal is 6, so we have $x = (176-6)/2 + 6$, and we get the same answer of 91. Or, generalizing
$$
x = [(N/2+1)\times (N/3+1) - (N/6+1)]/2 + (N/6+1)
$$
which will work for 300 as well. In general cases when $N$ is not divisible by 6 small adjustments need to be made.
