Find the number of positive integers n such that $\sqrt{n+\sqrt{n+\sqrt{n}}}<10$ for any finite number of square root signs. Find the number of positive integers $n$ such that $\sqrt{n+\sqrt{n+\sqrt{n}}}<10$ for any finite number of square root signs.
I know something with squaring and repeating, but what does "finite number of square root signs" mean?
 A: It seems that the question means to ask you, as mentioned by the user lulu, that: 
Find all natural numbers $n$ such that each of $\sqrt n,\sqrt{n+\sqrt n}, \sqrt{n+\sqrt{n+\sqrt n}},\dots$ is less than 10. 
Which is a round-about way of asking:
Find all natural numbers $n$ such that $\sqrt{n+\sqrt{n+\sqrt {n+\dots}}}<10$.
To solve it, we may use the standard operating procedure:
$$\text{Let }\sqrt{n+\sqrt{n+\sqrt {n+\dots}}}=x$$
$$\Rightarrow\sqrt{n+x}=x$$
$$\Rightarrow x^2-x-n=0$$
$$\Rightarrow x=\frac{1\pm\sqrt{1+4n}}{2}$$
And we are given that $x<10$. So:
$$\frac{1\pm\sqrt{1+4n}}{2}<10$$
$$\pm\sqrt{1+4n}<19$$
$$1+4n<361$$
$$n<90$$
But for an infinite sequence would $n=90$ not be less than $10$. Thus, for a finite sequence $n=90$ shouldn't be a problem.
Had we got something like $n<89.92$ then we would have concluded that max value $n$ can assume is $89$.
Thus, for your case, $n$ can be any value from $1$ to $90$.
A: As the number of square roots goes from $1$ to $\infty$, the nested radical will increase monotonically from $\sqrt n$ to a limit of $\frac{1+\sqrt{1+4n}}2$. This means

*

*if a choice of $n$ produces a limit greater than $10$, it is inadmissible since some finite number of square roots will also produce a result greater than $10$

*if a choice of $n$ produces a limit less than $10$, it is admissible; all finite nested radicals of its form will produce lesser results

Now
$$\frac{1+\sqrt{1+4n}}2<10\implies n<90$$
This is the tighter bound on $n$, since $\sqrt n<10\implies n<100$. $n=90$ is also admissible though since we deal with a finite number of square roots, not the limit. There are thus $90$ admissible integers.
