1. [20] For how many positive integers $n \leq 1000$ does the equation in real numbers
$$
x^{\lfloor x\rfloor}=n
$$
have a positive solution for $x$ ? (For a real number $x,\lfloor x\rfloor$ denotes the largest integer that is not greater than $x$.)
Answer:
$$
412
$$

Solution: If $\lfloor x\rfloor=0$, then $x^{\lfloor x\rfloor}=1=1^{\lfloor 1\rfloor}$, so we can safely ignore this case, as it does not introduce new solutions.
If $\lfloor x\rfloor=k$ for some $k>0, x \mapsto x^{\lfloor x\rfloor}=x^{k}$ is a continuous and increasing function on the interval $[k, k+1)$. Therefore, the $x^{\lfloor x\rfloor}$ can take on any value in $\left[k^{k},(k+1)^{k}\right)$ when $\lfloor x\rfloor=k$. Because $5^{4}<$ $1000<5^{5}$, it suffices to count the number of integers in the interval $\left[k^{k},(k+1)^{k}\right)$ for $1 \leq k \leq 4$.
Thus, the number of valid $n$ is simply
$$
\sum_{k=1}^{4}(k+1)^{k}-k^{k}=(2-1)+(9-4)+(64-27)+(625-256)=412 .
$$