2 digits?

We have to evaluate the given number $mod\quad 100$ . Applying Euler's Theorem as $(17,100)=1$ and $\phi (100)=40$ , we get that ${ 17 }^{ 40 }\equiv 1(mod\quad 100)$ , thus, ${ 17 }^{ 2016 }={ \left( { 17 }^{ 40 } \right) }^{ 50 }\cdot { 17 }^{ 16 }\equiv 1\cdot { 17 }^{ 16 }(mod\quad 100)$ . As ${ 17 }^{ 2 }=289\equiv -11(mod\quad 100)$ , then ${ 17 }^{ 16 }={ ({ 17 }^{ 2 }) }^{ 8 }\equiv {( -11) }^{ 8 }={ 11 }^{ 8 }(mod\quad 100)$ . Note that $11=10+1$ , thus, $11^{ 8 }={ (10+1) }^{ 8 }=\sum _{ k=0 }^{ 8 }{ \left( \begin{matrix} 8 \\ k \end{matrix} \right) { 10 }^{ 8-k }\cdot { 1 }^{ k } }$ , therefore, ${ 11 }^{ 8 }\equiv \left( \begin{matrix} 8 \\ 7 \end{matrix} \right) { 10 }^{ 1 }\cdot { 1 }^{ 7 }+\left( \begin{matrix} 8 \\ 8 \end{matrix} \right) { 1 }^{ 8 }=80+1=81(mod\quad 100)$ . Hence, the last two digits of the given number are $81$ .