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?

Using induction,
If $10^r=1+3^sk$ where $3\nmid k$
$$10^{3r}=(10^r)^3=(1+3^sk)^3=1+3^{s+1}k+3^{2s+1}k^2+3^{3s}k^3\equiv1\pmod{3^{s+1}}$$
for $s+1\le2s+1\iff  s\ge0,3s\ge s+1\iff 2s\ge1$
Now for the base case $r=1=3^0, s=2$
So, by using weak induction, $$10^{3^a}-1$$ is divisible by $3^{a+2}$ for $a\ge0$
