Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$ I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.
So that means that $\frac{10^a-1}{9}$ must be divisible by 3.
$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.
When is that last statement true?

You can use induction to prove
$$\frac{10^n-1}{9n} \tag{1}\label{eq1A}$$
is an integer when $n = 3^k$. The base case of $k = 0$ gives $3^k = 1$, with \eqref{eq1A} becoming $\frac{10-1}{9} = 1$. Next, assume \eqref{eq1A} is an integer for $k = m$ for some integer $m \ge 0$. Then, for $k = m + 1$,
$$\begin{equation}\begin{aligned}
\frac{10^{3^{m+1}} - 1}{9(3^{m+1})} & = \frac{\left(10^{3^{m}}\right)^3 - 1}{\left(9(3^{m})\right)3} \\
& = \frac{(10^{3^{m}} - 1)(10^{3^{2m}} + 10^{3^{m}} + 1)}{\left(9(3^{m})\right)3} \\
& = \left(\frac{10^{3^{m}} - 1}{9(3^{m})}\right)\left(\frac{10^{3^{2m}} + 10^{3^{m}} + 1}{3}\right)
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
The first factor above is an integer based on the induction hypothesis, i.e., \eqref{eq1A} being an integer for $n = 3^{m}$. For the second factor, note
$$10 \equiv 1 \pmod{3} \implies 10^{3^{2m}} \equiv 10^{3^{m}} \equiv 1 \pmod{3} \tag{3}\label{eq3A}$$
This means
$$10^{3^{2m}} + 10^{3^{m}} + 1 \equiv 3 \equiv 0 \pmod{3} \implies 3 \mid 10^{3^{2m}} + 10^{3^{m}} + 1 \tag{4}\label{eq4A}$$
i.e., the second factor is also an integer. With the product of $2$ integers being an integer, \eqref{eq2A} shows \eqref{eq1A} is an integer for $n = 3^{m+1}$. This proves by induction that \eqref{eq1A} is an integer for $n = 3^k$ for all integers $k \ge 0$.
