8...8d5...5 is divisible by 11
Solution 1:
A number is divisible by $11$ if and only if the alternating sum of its digits is divisible by $11$. So effectively we only care about the middle part of the number: $$ \underbrace{8\ldots 8}_{48}\,8X5\,\underbrace{5\ldots 5}_{48} $$ For which $X$ is the criterion (not) satisfied? ($8-X+5 \equiv 0 \pmod{11}$)