, 2016

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

Since $199 \equiv 1 \pmod{9}$ we have $199^n \equiv 1 \pmod{9}$, and so $S_n \equiv 199^n \equiv 1 \pmod{9}$ for $n=1, 2, \ldots$. Thus $S_n$ is among the numbers $1, 10, 19, 28, \ldots$ 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 $199 \equiv 1 \pmod{11}$ ($198 = 18 \cdot 11$), hence $199^n \equiv 1 \pmod{11}$; thus $199^n - 1$ is divisible by $11$. Let its digits at odd (respectively even) positions have sum $a$ (respectively $b$). Then $a \equiv b \pmod{11}$. On the other hand $a+b$ equals the digit sum of $199^n - 1$. Hence $0 \le a, b \le 9$, implying that $a \equiv b \pmod{11}$ is possible only if $a=b$. However then $2a=9$, which is a contradiction. The solution is complete.