Remainder when $2005^{2002} + 2002^{2005}$ is divided by $2003$

Question

Find the remainder when $2005^{2002} + 2002^{2005}$ is divided by $2003$ ?

Is there any better approach to this rather than using binomial theorem?

Answer 1

\begin{align*} 2005^{2002} + 2002^{2005} & \equiv (2)^{2002}+(-1)^{2005} \pmod{2003}\\ & \equiv 1-1 \pmod{2003}\\ & \equiv 0 \pmod{2003}. \end{align*} The second last step follows from Fermat's little theorem (since $2003$ is prime), we have $$2^{2002} \equiv 1 \pmod{2003}.$$