jmc

Notice that $3^{17}+3^{10}=3^{10}\cdot (3^7+1)$; thus $9$ divides into $3^{17}+3^{10}$. Furthermore, using the sum of seventh powers factorization, it follows that $3+1=4$ divides into $3^7+1$.

Using the divisibility criterion for $4$, we know that $\overline{AB}$ must be divisible by $4$. Thus $B$ is even and not divisible by $3$. Also, $A$ is odd, so $\overline{AB}=10A+B$, where $4$ does not divide into $10A$. Thus, $4$ cannot divide into $B$ either, otherwise $10A+B$ would not be divisible by $4$. Then, $B$ must be equal to $2$.

Using the divisibility criterion for $9$, it follows that $3(A+B+C)$ is divisible by $9$, that is $3$ divides into $A+C+2$. Thus, $A+C=4,7,10,13,16(\ast )$. Using the divisibility criterion for $11$, since $$\begin{array}{rl}10^8\cdot A+10^7\cdot B+\cdots +B& \equiv (-1{)}^8\cdot A+(-1{)}^7\cdot B+\cdots +B\\ & \equiv A-B+\cdots +B\\ & \equiv -1(\mathrm{mod}11),\end{array}$$then the alternating sum of digits, which works out to be $B+C-A\equiv -1(\mathrm{mod}11)$. Thus, $2+C-A$ is either equal to $10$ or $-1$, so $A-C=3,-8$.

In the former case when $A-C=3$, summing with $(\ast )$ yields that $2A\in \{7,10,13,16,19\}$, of which only $A=5$ fit the problem conditions. This yields that $C=2$. However, we know that $B$ and $C$ are distinct, so we can eliminate this possibility. Thus, $A-C=-8$, of which only $C=9,A=1$ works. The answer is $\boxed{129}$.