Positive ordered pair of $(10x)^2+(5y)^2<1,000,000$ How can I find the number of ordered pair $(x,y)\in\mathbb{Z}_{>0}$ where $(10x)^2+(5y)^2<1,000,000$?
Thanks
 A: To get an approximate answer, note that the ellipse with axes [-100,100] in $x$ and $[-200,200]$ in $y$ has area $20000\pi\approx 62864$.  This is the approximate number of lattice points in the ellipse.  $600$ are on the axes, and a quarter of what is left is in the first quadrant, so there are approximately $\frac 14 \cdot 62264=15566$ solutions
A: $(10x)^2+(5y)^2<1000000$
$(5y)^2<1000000-(10x)^2$
$5y<\sqrt{1000000-(10x)^2}$
$y<2\sqrt{10000-x^2}$
Let $x=1$ for an example $y<199.9899997$ or one can say that there are $199$ possible ordered pairs $(x,y)$ when $x=1$.
Now you need to workout $\displaystyle \sum_{x=1}^{99}\left\lfloor2\sqrt{10000-x^2}\right\rfloor$.
With some help from Wolfram, we get $\displaystyle \sum_{x=1}^{99}\left\lfloor2\sqrt{10000-x^2}\right\rfloor=15552$ .
A: Here is a brute force solution using Mathematica:

n = 0; For[x = 1, x < 100, x++, 
 For[y = 1, (10 x)^2 + (5 y)^2 < 1000000, y++, n++]]; n

returns
15548

A: I'm trying to answer this myself, check my work if I'm correct:
Since x can't be more than or equal to 100, and for y can't be more than or equal to 200, so there's 99*199 combinations possible?
