# Problem 4.

Find the number of numbers $N$ in the set $\{1,2, \ldots, 2018\}$ for which there exist positive solutions $x$ to the equation

$$
x^{[x]}=N
$$

( $[x]$ is the integer part of the real number $x$, i.e., the greatest integer not exceeding $x$)
# Solution.

Notice that suitable numbers $N$ for $x$ such that $[x]=n$ are the numbers from $n^{n}$ to $(n+1)^{n-1}$, that is, exactly the numbers for which $[\sqrt[n]{N}]=n$. Such numbers (among the numbers from 1 to 2018) are the number 1, the numbers from $2^{2}$ to $3^{2-1}$ (there are exactly 5 of them), the numbers from $3^{3}$ to $4^{3-1}$ (there are exactly 37 of them), and the numbers from $4^{4}$ to $5^{4-1}$ (there are exactly 369 of them).

In total, we get $1+5+37+369=412$ numbers.

Answer: 412 numbers.