Cute! Solved with CyclicISLscelesTrapezoid We claim the answer is yes. Note that the number $A$ and $\overline{(A+1)}11111111$ (with $8$ $1's$) are congruent modulo $9$. Hence, take $A$ and add $9$s until you get the number $\overline{(A+1)}11111111$, and then delete the $1$s to get $A+1$ $\quad \blacksquare$