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

$5^3=125=2\cdot61+3$, so $5^3\equiv 3\pmod{61}$. $3^5=243=4\cdot61-1$, so $5^{15}\equiv 3^5\equiv-1\pmod{61}$. Finally, $5^{20}=5^{15}\cdot5^3\cdot5^2\equiv-1\cdot3\cdot25\equiv-75\equiv-14\equiv47\pmod{61}$.
