The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0) is

We are asked to count all lattice points (points whose coordinates are both integers) that lie strictly inside the triangle whose vertices are $$A(0,0),\;B(0,41),\;C(41,0).$$

The most direct way is to use Pick’s Theorem, which applies to any simple polygon whose vertices have integer coordinates. The theorem states

$$\text{Area}=I+\frac{B}{2}-1,$$

where

$$I=\text{number of interior lattice points},\qquad B=\text{number of lattice points on the boundary}.$$

We first evaluate the area of the triangle. Since the triangle is right-angled at the origin, the area is half the product of the perpendicular sides:

$$\text{Area}=\frac12\times 41\times 41=\frac{1681}{2}=840.5.$$

Next we must count $$B,$$ the lattice points lying on the three sides.

1. On the vertical side $$AB$$ from $$(0,0)$$ to $$(0,41)$$, the $$x$$-coordinate is always $$0,$$ while $$y$$ runs through all integers from $$0$$ to $$41.$$ Thus the number of lattice points on $$AB$$ is

$$41-0+1=42.$$

2. On the horizontal side $$AC$$ from $$(0,0)$$ to $$(41,0),$$ the $$y$$-coordinate is always $$0,$$ while $$x$$ runs from $$0$$ to $$41.$$ Therefore the side $$AC$$ also contributes

$$41-0+1=42$$

lattice points.

3. For the hypotenuse $$BC,$$ join $$(0,41)$$ to $$(41,0).$$ Its equation is $$x+y=41.$$ We need all integer solutions with $$x\ge 0$$ and $$y\ge 0.$$ If $$x$$ takes the values $$0,1,2,\dots,41,$$ then $$y$$ is automatically $$41-x,$$ remaining an integer in the permitted range. Hence this side also possesses

$$41-0+1=42$$

lattice points.

Adding the counts of all three sides gives $$42+42+42=126,$$ but each of the three vertices $$(0,0),\;(0,41),\;(41,0)$$ has just been tallied twice, once for each adjoining side. To correct this double counting we subtract the number of vertices once:

$$B=126-3=123.$$

Now substitute the known area and boundary count into Pick’s formula:

$$840.5=I+\frac{123}{2}-1.$$

Simplifying the right-hand side step by step, we have

$$\frac{123}{2}=61.5,$$

so

$$I+61.5-1=I+60.5.$$

Equating the two sides gives

$$I+60.5=840.5.$$

Subtract $$60.5$$ from both sides:

$$I=840.5-60.5=780.$$

Thus exactly $$780$$ lattice points lie strictly inside the given triangle.

Hence, the correct answer is Option A.