South African Mathematical Olympiad, Third Round

The number $N = \underline{ababa}$ is divisible by 11 if and only if $a-b+a-b+a = 3a-2b$ is divisible by 11. To find the smallest possible $N$ with this property, we observe that $a=1$ (the smallest possible value for $a$) and $b=7$ offer a solution: $3 \cdot 1 - 2 \cdot 7 = -11$ is divisible by 11 (and no smaller $b \in \{0, 1, \dots, 9\}$ works). So $N = 17171$ is the smallest number of the required form that is divisible by 11. Similarly, the largest possible value for $a$ is $a=9$, and we do have a solution in this case, with $b=8$: $3 \cdot 9 - 2 \cdot 8 = 11$ is divisible by 11 (and no larger $b \in \{0, 1, \dots, 9\}$ works). So $N = 98989$ is the largest number of the required form that is divisible by 11.