Simplest method to find $5^{20}$ modulo $61$ What is the simplest method to go about finding the remainder of $5^{20}$ divided by $61$?

Note that $5^3=125 = 3\pmod{61}$, so $5^{20}=5^{18}.25=(5^3)^6.25=3^6.25\pmod {61}$.Now $3^5=243\pmod {61}=-1\pmod {61}$ $\implies$ $3^6.25\pmod {61}= -3.25\pmod{61}=-75\pmod{61}=47\pmod {61}$.
