A number theory problem by Wildan Bagus Wicaksono

Based on

${ (a+b) }^{ n }=\left( \begin{matrix} n \\ 0 \end{matrix} \right) { a }^{ n }{ b }^{ 0 }+\left( \begin{matrix} n \\ 1 \end{matrix} \right) { a }^{ n-1 }{ b }^{ 1 }+\left( \begin{matrix} n \\ 2 \end{matrix} \right) { a }^{ n-2 }{ b }^{ 2 }+...+\left( \begin{matrix} n \\ n-1 \end{matrix} \right) { a }^{ 1 }{ b }^{ n-1 }+\left( \begin{matrix} n \\ n \end{matrix} \right) { a }^{ 0 }{ b }^{ n }$

${ (1+1) }^{ 2017 }=\left( \begin{matrix} 2017 \\ 0 \end{matrix} \right) { 1 }^{ 2017 }{ 1 }^{ 0 }+\left( \begin{matrix} 2017 \\ 1 \end{matrix} \right) { 1 }^{ 2016 }{ 1 }^{ 1 }+\left( \begin{matrix} 2017 \\ 2 \end{matrix} \right) { 1 }^{ 2015 }{ 1 }^{ 2 }+...+\left( \begin{matrix} 2017 \\ 2016 \end{matrix} \right) { 1 }^{ 1 }{ 1 }^{ 2016 }+\left( \begin{matrix} 2017 \\ 2017 \end{matrix} \right) { 1 }^{ 0 }{ 1 }^{ 2017 }\\ { 2 }^{ 2017 }=\left( \begin{matrix} 2017 \\ 0 \end{matrix} \right) +\left( \begin{matrix} 2017 \\ 1 \end{matrix} \right) +\left( \begin{matrix} 2017 \\ 2 \end{matrix} \right) +...+\left( \begin{matrix} 2017 \\ 2016 \end{matrix} \right) +\left( \begin{matrix} 2017 \\ 2017 \end{matrix} \right)$

Notice the pattern below.

${ 2 }^{ 1 }\equiv 2(mod10)\\ { 2 }^{ 2 }\equiv 4(mod10)\\ { 2 }^{ 3 }\equiv 8(mod10)\\ { 2 }^{ 4 }\equiv 6(mod10)\\ { 2 }^{ 5 }\equiv 2(mod10)$

The pattern repeats after 4 patterns.

$2017 \equiv 1 (mod 4)$ .

Thus, the unit number of $2 ^{2017}$ equals the unit number of $2 ^ 1$ , is $2$ .