2009/10 AHSMC Part 1 Problem 13

With one square root sign, $\sqrt{n} < 10$ is equivalent to $n < 100$. With two square root signs, $\sqrt{n + \sqrt{n}} < 10$ is equivalent to $n + \sqrt{n} < 100$, which is in turn equivalent to $n < 91$. With three square root signs and $n < 91$, we have $\sqrt{n + \sqrt{n + \sqrt{n}}} < \sqrt{n + 10} \le 10$. With more square root signs, the same inequality will hold. There are $90$ positive integers which satisfy $n < 91$. The answer is (B).