Minimum value of digits sum

Careful, $10^{2002} \equiv 1 \pmod{2003}$ , which means that $10^{2002} + 2$ is not divisible by $2003$ .

By condering the Legendre symbol $\left(\frac{-2}{2003}\right) = \left(\frac{-1}{2003}\right)\left(\frac{2}{2003}\right) = -1 \times -1 = 1$ , we see that $-2$ is a quadratic residue modulo $2003$ . We also note that $\left(\frac{10}{2003}\right) = \left(\frac{2}{2003}\right)\left(\frac{5}{2003}\right) = -\left(\frac{2003}{5}\right) = -\left(\frac{3}{5}\right) = 1$ and so $10$ is a quadratic residue modulo $2003$ .

It is easy to check that $10$ has order $1001$ in the cyclic group $\mathbb{Z}_{2003}^\times \equiv C_{2002}$ , and so $10 \equiv x^2$ for some $x \in \mathbb{Z}_{2003}^\times$ , and it is clear that $x$ has order $2002$ , and so is a generator of the group. Since $-2$ is a quadratic residue modulo $2003$ it follows that $-2 \equiv y^2$ for some $y$ , and we will have $y \equiv x^u$ for some integer $u$ . Thus $-2 \equiv x^{2u} \equiv 10^u$ , and so there must exist some integer $u$ such that $10^u + 2$ is divisible by $2003$ . We do not need to find the actual value of $u$ to solve the problem! I think I found $u=301$ by computer search (it was a year ago).