The number of points,having both coordinates | vedclass

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Question

(A) The vertices of the triangle are $(0,0)$,$(41,0)$,and $(0,41)$.
The equation of the line passing through $(41,0)$ and $(0,41)$ is $x + y = 41$.

Vedclass · Accepted Answer

$780$

Vedclass · Answer

(A) The vertices of the triangle are $(0,0)$,$(41,0)$,and $(0,41)$.
The equation of the line passing through $(41,0)$ and $(0,41)$ is $x + y = 41$.
We need to find the number of integer pairs $(x, y)$ such that $x > 0$,$y > 0$,and $x + y < 41$.
For a fixed $x$,the possible values of $y$ are $1, 2, 3, dots, 40-x$.
The number of such points for a given $x$ is $40-x$.
Summing these values for $x = 1$ to $39$:
$N = \sum_{x=1}^{39} (40-x) = 39 + 38 + 37 + dots + 1$.
This is an arithmetic progression with $n = 39$ terms.
$N = \frac{n(n+1)}{2} = \frac{39 	imes 40}{2} = 39 	imes 20 = 780$.

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:

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