Let $a = 31$. Consider the set of integers $T = \{a, 8a, 8^{2}a, 8^{3}a, \cdots \}$. Does $T$ contain the integer:
$999999999900000000000090909090000000000000000008$?
So far I've deduced that if we work mod $9$ that the set $T$ can be reduced to a reduced residue system modulo $9$ by using $8$ as a root. Additionally, because $(a, 9) = (31, 9) = 1$ we can eliminate $a$ from the elements of $T$.
Continuing, remark that $ord_{9}(2) = \phi(9) = 6$ and $ord_{9}(2^{3}) = \frac{6}{3} = 2$. Taking $8^{k} \mod 9$ for $k \in \mathbb{Z^{+}}$ we see that we get the set of reduced residues $\{8, 1\}$. $8$ is an element of this set so $T$ contains $999\ldots 0008$.
That's where I am so far but I have a feeling that I went wrong early on in this one.