45th 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 $$\{\begin{array}{l}n\equiv 1(\mathrm{mod}p-1)\\ n\equiv -10(\mathrm{mod}p)\end{array}.$$ Then by Fermat's theorem, $$10^n+n\equiv 10+n\equiv 10-10=0(\mathrm{mod}p)$$ and $$x^n+n\equiv x+n\equiv x-10\not\equiv 0(\mathrm{mod}p).$$ It follows that $p$ divides $10^n+n$ but does not divide $x^n+n$, $x\ne 10$, a contradiction. Hence $x=10$ only.