Here is a similar example: Suppose you want to find $x$ where ( $0\le x\le 100$) and ${17}^{13}\equiv xmod101$. As ${17}^{13}$ is too large for most calculators to show exactly we start with ${17}^6=24137569$ and, first dividing this by 101, we find that ${17}^6=(238985)(101)+84$ so we now know that ${17}^6\equiv 84mod101.$

The next step is to use this to tackle ${17}^{13}$.

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{17}^{13}}& =({17}^6{)}^2\times 17\\ \multicolumn{1}{c}{}& \equiv {84}^2\times 17\equiv 119952mod101\\ \multicolumn{1}{c}{119952}& =1187\times 101+65\\ \multicolumn{1}{c}{}& \equiv 65mod101.\end{array}}$

Hence $x=65$.