Find the largest integer $$𝑛$$ such that a square of side length $$𝑛$$ is contained in a circular disc of area $$1000$$.

The circular disc has area $$1000$$, so

$$\pi r^{2}=1000 \; \Longrightarrow \; r=\sqrt{\frac{1000}{\pi}}.$$

The diameter of the disc is therefore

$$d=2r = 2\sqrt{\frac{1000}{\pi}}.$$

For the largest possible square to be fully contained in the disc, its four vertices must lie on the circle. In that position the diagonal of the square equals the diameter of the circle.

If $$n$$ is the side length of the square, its diagonal is $$n\sqrt{2}$$. Setting this equal to the diameter gives

$$n\sqrt{2}=2\sqrt{\frac{1000}{\pi}} \quad\Longrightarrow\quad n=\frac{2}{\sqrt{2}}\sqrt{\frac{1000}{\pi}} =\sqrt{\frac{2000}{\pi}}.$$

Numerically,

$$\sqrt{\frac{2000}{\pi}}\approx\sqrt{636.62}\approx 25.24.$$

Because $$n$$ must be an integer no larger than this value, the greatest possible integer $$n$$ is $$25$$.

Answer: 25