Sum of remaining of 2^k divided by 2003

Question

Sorry for my interruption. I am looking for an answer to this question: Calculating $$\sum_{k=1}^{2002}r_k$$ in which $r_k$ is the remaining of $2^k$ divided by $2003$. I thought that $2$ was a primitive root of $2003$, but I was wrong, in fact $ord_{2003}(2) = 286$. I hope you can help me on this question.

Answer 1

Hint: Use geometric series for $2^k$.

Solution: (Hover to reveal)

$$\sum_{k=1}^{2002}{r_k}=\left(\sum_{k=1}^{2002}{2^k}\right) \pmod{2003}$$ $$=\frac{2(1-2^{2002})}{-1}=2^{2003}-2$$ By the geometric series formula. Note that 2003 is prime so by Fermat's Little Theorem, $2^{2003}-2 \equiv 0 \pmod{2003}$