If $r$ is the remainder of dividing the number $17*2^{635}$ by $33$, then $8$ divides $r$

Question

How can I prove the statement below?

$r\in \mathbb{Z}$ such that: $ 0\leq r\leq 32$

If r is the remainder of dividing the number $17*2^{635}$ by 33, then 8 divides r

Answer 1

We can calculate $r$ explicity.

Notice that $2^a\bmod 33$ is periodic with period $lcm(11-1,3-1)=10$.

Hence we need only calculate $2^5\bmod 33$ which is $-1$. So $r$ is $-17\equiv 16$, and $8$ divides $16$

Answer 2

$\!\!\bmod \color{#c00}{2^{\large 5}\!+1}\!:\,\ \color{#c00}{2^{\large 5}\equiv -1}\,\Rightarrow\, 17(\color{#c00}{2^{\large 5}})^{\large 127}\!\equiv 17(\color{#c00}{-1})^{\large 127}\!\equiv -17\equiv 16\,$ via Congruence Power Rule.