6. Find the smallest integer $n$ such that $\sqrt{n+99}-\sqrt{n}<1$.
Solution: 2402
This is equivalent to
$$
\begin{aligned}
\sqrt{n+99} & <\sqrt{n}+1 \\
n+99 & <n+1+2 \sqrt{n} \\
49 & <\sqrt{n}
\end{aligned}
$$

So the smallest integer $n$ with this property is $49^{2}+1=2402$.