23. Determine the largest prime factor of the sum $\sum_{k=1}^{11} k^{5}$.
23. Answer: 263
We consider the sum $\sum_{k=1}^{11}\left(6 k^{5}+2 k^{3}\right)$ and observe that
$$
\begin{array}{l}
\sum_{k=1}^{11}\left(6 k^{5}+2 k^{3}\right) \\
=\sum_{k=1}^{11}\left\{k^{3}(k+1)^{3}-k^{3}(k-1)^{3}\right\} \\
=11^{3} \times 12^{3} .
\end{array}
$$

Hence
$$
\begin{array}{l}
\sum_{k=1}^{11} k^{5} \\
=\frac{11^{3} \times 12^{3}}{6}-\frac{2}{6} \sum_{k=1}^{11} k^{3} \\
=2 \times 11^{3} \times 12^{2}-\frac{1}{3}\left(\frac{11^{2} \times 12^{2}}{4}\right) \\
\left.=11^{2} \times 12 \times(2 \times 11 \times 12-1)\right) \\
=11^{2} \times 12 \times 263 .
\end{array}
$$

Finally. since $\sqrt{256}=16$ and the numbers $2,3,5,7,11$ and 13 do not divide 263 , we conclude that 263 is the largest prime factor of $\sum_{k=1}^{11} k^{5}$.