Argentine National Olympiad 2016

The minimum of $S_n$ is $19$, attained already for $n=1$.

Since $199\equiv 1(\mathrm{mod}9)$ we have $199^n\equiv 1(\mathrm{mod}9)$, and so $S_n\equiv 199^n\equiv 1(\mathrm{mod}9)$ for $n=1,2,\dots$. Thus $S_n$ is among the numbers $1,10,19,28,\dots$ Clearly $S_n=1$ never holds, so to prove $\min S_n=19$ it suffices to show that $S_n=10$ is also impossible.

Suppose on the contrary that $S_n=10$ holds for some $n$. Consider the number $199^n-1$. Because $199^n$ ends in $1$ or $9$, the digit sum of $199^n-1$ is $S_n-1=10-1=9$. Now observe that $199\equiv 1(\mathrm{mod}11)$ ($198=18\cdot 11$), hence $199^n\equiv 1(\mathrm{mod}11)$; thus $199^n-1$ is divisible by $11$. Let its digits at odd (respectively even) positions have sum $a$ (respectively $b$). Then $a\equiv b(\mathrm{mod}11)$. On the other hand $a+b$ equals the digit sum of $199^n-1$. Hence $0\le a,b\le 9$, implying that $a\equiv b(\mathrm{mod}11)$ is possible only if $a=b$. However then $2a=9$, which is a contradiction. The solution is complete.