Show that $3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other. 
$3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other.  Prove it!

$3^{-1}=\overline{0\color{red}{1}}_2$
$3^{-2}=\overline{000\color{red}{111}}_2$
$3^{-3}=\overline{000010010\color{red}{111101101}}_2$
$\ldots$
These are the binary representations - add a minus sign to the left hand side and the right hand side is the 2-adic representation.

Lemma: For any sequence of $k$ digits $A$, where $B$ is the opposite sequence of digits: $$0.\overline{AB}_2 = \frac{A + 1}{2^k + 1}.$$
Proof: Let $x = 0.\overline{AB}$. Then
$$
2^{k} x = A.\overline{BA}.
$$
Therefore,
$$
x + 2^k x = A.\overline{11\ldots 1} = A + 1 \quad \implies \quad x = \frac{A + 1}{2^k + 1}.
$$

So we have to show that $3^{-n} = \frac{1}{3^n}$ has this form, $\frac{A+1}{2^k + 1}$, where $A$ is between $0$ and $2^{k} - 1$. Because $3^{-n}$ is definitely between $0$ and $1$, this is equivalent to just finding $k$ such that $3^n \mid 2^{k} + 1$.
From the pattern in the examples in your question, we guess that we should pick
$$
k = 3^{n-1}.
$$
So we show by induction that
$$
3^n \mid 2^{3^{n-1}} + 1.
$$
Base case: $3^1 \mid 2 + 1 = 3$.
Inductive step:
Let's assume that $3^n \mid 2^{3^{n-1}} + 1$, for some $n$. Specifically, let
$2^{3^{n-1}} = a \cdot 3^n - 1$.
 Then
\begin{align*}
2^{3^n} + 1
  &= \left(2^{3^{n-1}}\right)^3 + 1 \\
  &= \left(a \cdot 3^n - 1\right)^3 + 1 \\
  &= a^3 3^{3n} - 3 a^2 3^{2n} + 3 a 3^n - 1 + 1 \\
  &\equiv 0 - 0 + 0 - 1 + 1 \pmod{3^{n+1}} \\
  &= 0.
\end{align*}
So $2^{3^n} + 1$ is divisible by $3^{n+1}$, and the induction is complete.
