23. Find the smallest positive integer $k$ such that $(k-10)^{5026} \geq 2013^{2013}$.
23. Answer: 55
$(k-10)^{4026}=\left((k-10)^{2}\right)^{2013} \geq 2013^{2013}$ is equivalent to to $(k-10)^{2} \geq 2013$. As $k-10$ is an integer and $44^{2}<2013<45^{2}$, the minimum value of $k-10$ is 45 , and thus the minimum value of $k$ is 55 .