If the positive integer $$m$$ is such that $$\frac{m}{2006}$$ is closest to $$\sqrt{2}$$, then $$m =$$ ________

We need the integer nearest to $$2006\sqrt{2}$$. We have $$\left(2006\sqrt{2}\right)^2 = 2 \times 4024036 = 8048072$$, while $$2836.5^2 = 8045732.25$$ and $$2837.5^2 = 8051406.25$$. Hence $$2836.5 < 2006\sqrt{2} < 2837.5$$, and the required integer is $$m = 2837$$ (indeed $$2006\sqrt{2} \approx 2836.91$$).