Sum of digits of the multiple of 2003

Question

Can you find the minimum value of $S(2003n)$, with $S(x)$ being the sum of digits in decimal form and n being any positive integer? (For example, $S(2018)=2+0+1+8=11$ )

Answer 1

Direct computation shows that the order of $10\pmod {2003}$ is $1001$. Note that this is odd.

Claim $1$: $1$ is impossible.

Pf: The only numbers with digit sum $1$ are powers of $10$ and none of those are divisible by $2003$.

Claim $2$: $2$ is impossible.

Pf: The only numbers with digit sum $2$ are those of the form $10^a(10^b+1)$ for $a,b≥0$. If $b=0$, we have the numbers of the form $2\times 10^n$, none of which are divisible by $2003$, so we may assume that $b>0$. In order for such a number to be divisible by $2003$ we'd need $10^b\equiv -1\pmod {2003}$. But this would imply that $10^{2b}\equiv 1\pmod {2003}$, whence that $1001\,|\,2b$ which implies that $1001\,|\,b$. Taking a minimal possible $b$ shows that this is impossible.

Claim $3$: $3$ is possible

Pf: By direct computation, $10^{301}\equiv -2\pmod {2003}$.

Hence $3$ is the minimum.