National Olympiad of Argentina

The answer is $5$ for all $n\ge 1000$. If $n\ge 200$ we have $n^2+n+200<n^2+2n+1=(n+1{)}^2$, so $n<\sqrt{n^2+n+200}<n+1$ and $\lfloor \sqrt{n^2+n+200}\rfloor =n$. It also follows that $\sqrt{n^2+n+200}$ is not an integer, moreover $\sqrt{n^2+n+200}$ is irrational. Let $k$ be the first digit of $\sqrt{n^2+n+200}$ after the decimal point, $0\le k\le 9$. Then $n+\frac{k}{10}<\sqrt{n^2+n+200}<n+\frac{k+1}{10}$, or

$$10n+k<10\sqrt{n^2+n+200}<10n+(k+1).$$

The inequalities are strict as $\sqrt{n^2+n+200}$ is irrational. Squaring and simplification gives $$20nk+k^2<100n+20000<20n(k+1)+(k+1{)}^2.$$

The left inequality implies $20n(k-5)<20000$, $n(k-5)<1000$. Given $n\ge 1000$, we see that $k\le 5$. Otherwise $k-5\ge 1$ and $n(k-5)\ge n\ge 1000$. The right inequality can be rewritten as $20000<20n(k-4)+(k+1{)}^2$. Hence $20000<20n(k-4)+10^2$ because $k\le 9$; thus $1000<n(k-4)+5$. So $n(k-4)>0$ which implies $k\ge 5$. Now $k\le 5$ and $k\ge 5$ lead to $k=5$.