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 $5^3 = 125 = 2(61)+3$. So $5^{20} = (5^3)^6\times 5^2\equiv 3^6\times 5^2 \pmod{61}$.
Now, $3^3\equiv 27$, $3^3\times 5 = 135 \equiv 13\pmod{61}$. So
$$5^{20}\equiv (5^3)^6\times 5^2\equiv 3^6\times 5^2 =(3^3\times 5)^2 \equiv 13^2 \equiv 169\pmod{61}$$
and since $169 = 2(61) + 47$, we finally have that $5^{20}\equiv 47\pmod{61}$. 
If you want to be systematic, then repeated squaring works well: $20 = 16 + 4$. 
So first compute $5^2=25$. Then square that and reduce modulo $61$ to get $5^4$; then square that and reduce to get $5^8$; square again to get $5^{16}$. Then multiply the results of $5^4$ and $5^{16}$ to get $5^{20}$. This amounts five products and reductions.
