Let $\displaystyle m$ be a natural number. If $\displaystyle m^{53}$ has remainder $\displaystyle 31$ when divided by $\displaystyle 143$, what is the remainder of $\displaystyle m$ when divided by $\displaystyle 143$ ?
$n=143=11\cdot13$. So $\phi(n)=10\cdot12=120$. Now 53 is prime to 120 and so has an inverse mod 120. With some computer help, $1=19\cdot120+(-43)\cdot53$. So the inverse of 53 mod 120 is 120-43=77. Hence $m\equiv m^{53\cdot 77}\equiv 31^{77}\equiv 70 (mod\,\, 143)$. Again, the computer helped.