How common are integer solutions to this problem? Consider the problem of finding the smallest number $x \in \mathbb{R}$ that satisfies the constraint $x \lfloor x \rfloor \geq S$ where $S$ is an integer (of course, $\lfloor x \rfloor$ simply means $x$ rounded down).
Often, solving this problem will yield an integer solution for $x$. For example, if $S = 100$, then the solution is $x = 10$ (since $x \lfloor x \rfloor = 10\lfloor 10 \rfloor = 10^2 \geq 100$). Similarly, if $S = 99$, then the solution remains at $x = 10$ (clearly, $x = 10$ satisfies the inequality; and if $x < 10$, then $x \lfloor x \rfloor < 10\cdot9 = 90 < 99$ so the inequality is violated). But for other values of $S$, e.g. $S = 101$, the solution is not an integer (it is $x = 10.1$.)
Question: Suppose we solve this problem for every possible value $S = 0, 1, 2, ...$, each time updating the proportion of solutions $p$ that are integer solutions. Is there some way to evaluate the limit of $p$ (as $S \rightarrow \infty$)?
 A: Let $S\geq0$ and let $n:=\lfloor\sqrt{S}\rfloor$ so that
$$n^2\leq S<(n+1)^2.$$
If $S<n(n+1)$ then $x=\frac{S}{n}$ satisfies $\lfloor x\rfloor =n$ and hence $x\lfloor x\rfloor=S$.
If $S\geq n(n+1)$ then $x=n+1$ is the smallest real number satisfying $x\lfloor x\rfloor\geq S$.
In particular this shows that $x$ is an integer if and only if
$$n(n+1)\leq S\leq(n+1)^2,$$ for some positive integer $n$. From here it is not difficult to show that the limit equals $\tfrac12$.
A: If $n(n-1)\leq S\leq n^2$ for some integer $n\geq 2$ then the answer is an integer (in fact, is the integer $n$). This is because for any $x<n$ we have $\lfloor x\rfloor\leq n-1$ and so $x\lfloor x\rfloor< n(n-1)$.
Conversely, if $(n-1)^2<S<n(n-1)$ the answer is not an integer, since $\lfloor S/(n-1)\rfloor=n-1$ so $x=S/(n-1)$ works (but $n-1$ doesn't).
There are $2n-1$ values of $S$ with $(n-1)^2<S\leq n^2$ for each $n$. Of these, $n+1$ work and $n-2$ don't. These intervals cover all natural numbers. So the limit will be $1/2$.
