Use Fermat's Little Theorem to prove that
$11|(9n^{23}-5n^{13}+7n^3)$
Use Fermat's Little Theorem to prove that
$11|(9n^{23}-5n^{13}+7n^3)$
$n^{23}\equiv n^{13}\equiv n^3\pmod{11}$ so we get that $ 9n^{23}-5n^{13}+7n^3\equiv11n^3\equiv0\pmod{11} $ Thus, $11|(9n^{23}-5n^{13}+7n^3)$.
If modular arithmetic is unfamiliar then one may use that $\rm\:11\mid \color{#C00}{n-n^{11}}\:$ by little Fermat, hence
$\rm 11\mid7n^3\! - 5n^{13}\! + 9 n^{23}\, =\ (7n^2\! + 2n^{12}) (\color{#C00}{n-n^{11}}) + 11 n^{23}$