Number of integral coordinates in a given region. The number of points, having both coordinates as integers, that lie in the interior of the triangle with
vertices $(0,0) ,(0, 41$) and $(41,0)$ , is:
(1) 901   (2) 861   (3) 820   (4) 780.
I tried to find out the number of points manually, but it didn't look like the optimal method.
Thanks in advance.
 A: Manually seems fine to me. First draw the triangle.
Consider the integral points on the line $x=1$. The point on the hypotenuse is $(1,40)$. The number of points inside the triangle are $39$, that is, $\{1,2\dots,39\}$.
Similarly number of integral points on $x=2$ are $38$, that is $\{1,2,\dots,38\}$.
So basically you have to sum up $39+38+\dots+1 = \frac{40.39}{2}=780.$
A: The problem can be approached by finding the area of the triangle in two ways:
First, as $\displaystyle A = \frac{1}{2}(41^2) = \frac{1681}{2}$
Second, by an application of Pick's theorem $\displaystyle A = i + \frac{b}{2} - 1$
Now boundary points $b$ can be calculated by considering that there is a single origin point $(0,0)$ common to both cathetuses, $40$ points unique to each of the two cathetuses, and $42$ points along the hypotenuse (inclusive of the end points where it intersects with the cathetuses).
So $\displaystyle b = 1+ 2(40)+42 = 123$, giving $\displaystyle A = i + \frac{121}{2}$
Equating the two expressions for $A$, we get the number of interior points $i = 780$.
