Mongolian Mathematical Olympiad

Only $x = 10$.

Assume the contrary and a prime $p$ that does not divide $x - 10$. By the Chinese Remainder Theorem we can find a positive integer $n$ such that
$$\begin{cases} n \equiv 1 \pmod{p-1} \\ n \equiv -10 \pmod{p} \end{cases}.$$
Then by Fermat's theorem,
$$10^n + n \equiv 10 + n \equiv 10 - 10 = 0 \pmod{p}$$
and
$$x^n + n \equiv x + n \equiv x - 10 \not\equiv 0 \pmod{p}.$$
It follows that $p$ divides $10^n + n$ but does not divide $x^n + n$, $x \neq 10$, a contradiction. Hence $x = 10$ only.