Let S be the set of all triangles in the xy plane, each having one vertex at the origin and the other two vertices lie on the coordinate axes with integral coordinates. If each triangle in S has area 50sq unit, then number of elements in set S is?
Let the coordinates of the other two vertices be $(a,0)$ and $(0,b)$
Clearly $$\frac {|a||b|}{2}=50$$ $$|a||b|=100$$
Let’s consider both $a,b$ as positive for now
$$100=2^2\times 5^2$$
And consider an order pair $(a,b)$
So number of such order pairs will be $$\frac{4!}{2!2!}$$ $$=6$$
This was first quadrant. So for 4 quadrants, answer will be $24$, but given is 36. Which combinations am I missing?
Note: manually counting cases gives 9 possibilities, which gives the right answer, but I want to use a formula to solve
$100 = 2^2 \cdot 5^2$.
So possible integer values of $a$ comes from three choices for factor $2$ - either factor $2$ is not present, multiplied one time or multiplied two times $(2^0, 2^1$ or $ 2^2)$. We have same three choices for $5$. We also observe as we choose $a$, $b$ is automatically decided and so we obtain ordered pair $(a, b)$.
That gives us total of $3\cdot 3 = 9$ combinations in one quadrant. We multiply by $4$ to obtain $36$ possible combinations in total across all $4$ quadrants.