10. Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$. How many positive integers less than 2005 can be expressed in the form $\lfloor x\lfloor x\rfloor\rfloor$ for some positive real $x$ ?
Solution:
990

Let $\{x\}=x-\lfloor x\rfloor$ be the fractional part of $x$. Note that
$$
\lfloor x\lfloor x\rfloor\rfloor=\lfloor(\lfloor x\rfloor+\{x\})\lfloor x\rfloor\rfloor=\lfloor x\rfloor^{2}+\lfloor\{x\}\lfloor x\rfloor\rfloor .
$$

Because $\{x\}$ may take on any value in the half-open interval $[0,1)$, the quantity $\lfloor\{x\}\lfloor x\rfloor\rfloor$ can take on any integer value between 0 and $\lfloor x\rfloor-1$, inclusive.
If $\lfloor x\rfloor=n$, then $\lfloor x\lfloor x\rfloor\rfloor$ can be any of the numbers $n^{2}, n^{2}+1, \ldots, n^{2}+n-1$. In other words, there are precisely $n$ possible values that $\lfloor x\lfloor x\rfloor\rfloor$ can take, and moreover, all of them are less than $(n+1)^{2}$. Because $44^{2}+43=19792005$, $n$ can range between 1 and 44 , inclusive. Therefore, the answer is
$$
\sum_{n=1}^{44} n=\frac{44 \cdot 45}{2}=990 .
$$