Modular exponentitation (Finding the remainder) I want to find the remainder of $8^{119}$ divided by $20$ and as for as i do is follows:
$8^2=64\equiv 4 \pmod {20} \\
8^4\equiv 16 \pmod {20} \\
8^8\equiv 16 \pmod {20}\\
8^{16}\equiv 16 \pmod {20}$
from this i see the pattern as follows $8^{4\cdot 2^{n-1}} \text{is always} \equiv 16 \pmod {20} \,\forall n \ge 1$
So,
$\begin{aligned} 8^{64}.8^{32}.8^{16}.8^7 &\equiv 16.8^7 \pmod{20}\\
&\equiv 16.8^4.8^3 \pmod{20} \\
&\equiv 16.8^3 \pmod {20}\end{aligned}$
And i'm stuck. Actually i've checked in to calculator and i got the answer that the remainder is $12$. But i'm not satisfied cz i have to calculate $16.8^3$
Is there any other way to solve this without calculator. I mean consider my condisition if i'm not allowed to use calculator.
Thanks and i will appreciate the answer.

Using $\, ab\bmod ac\, =\, a\,(b\bmod c) $ = mod Distributive Law to factor out $\,a =4\,$ yields
$\ \ \ \ 8^{\large 3+4K}\!\bmod 20\, =\, 4\,(\underbrace{2\cdot \color{#0a0}{8^{\large 2}}\,\color{#c00}8^{\large \color{#c00}{4}K}}_{\Large 2\,   (\color{#0a0}{-1})\,\color{#c00}1^{\Large K}}\bmod 5) = 4(3)\ $ 
