What is the remainder of dividing $(116+17^{17})^{21}$ by $8$?

Question

What is the remainder of dividing $(116+17^{17})^{21}$ by $8$?

How to solve this? Solving the congruence (find the value of $a$) in $$(116+17^{17})^{21}\equiv a\pmod8$$ and how is this done?

Answer 1

As $17\equiv1\pmod 8,17^{17}\equiv1\pmod 8$ and $116\equiv4\pmod 8$

$\displaystyle\implies 116+17^{17}\equiv 4+1\pmod8\equiv5$

$\displaystyle\implies(116+17^{17})^{21}\equiv 5^{21}\pmod8 $

Now, $\displaystyle5^2=25\equiv1\pmod 8\implies 5^{21}=5\cdot(5^2)^{10}\equiv5\cdot1^{10}\pmod 8\equiv5$

Answer 2

Hint: The stuff inside the parentheses is odd. Any odd number squared is $\equiv 1 \pmod 8$