<s>[INST] 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?
 [/INST] The statement is equivalent to proving $$10^{3^k}-1\equiv_{3^{k+2}}0$$
Notice that $\phi(3^{k+2})=3^{k+2}-3^{k+1}=2\cdot 3^{k+1}$, so that for a unit $a$ we have $$a^{2\cdot 3^{k+1}}\equiv_{3^{k+2}}1$$ We also have

Claim: $$10^{3^k}\equiv_{3^{k+2}}10^{3^{k+1}}$$

Proof: $$10^{3^k}\equiv_{3^{k+2}}10^{3^{k+1}-2\cdot 3^{k}}\equiv_{3^{k+2}}10^{3^{k+1}}$$
Let $t=10^{3^k}$, then the above claim may be written as $$t^3\equiv_{3^{k+2}}t\iff t(t+1)(t-1)\equiv_{3^{k+2}}0$$ Both $t,t+1$ are units since $t,t+1\not\equiv_{3}0$, therefore we must have $t-1\equiv_{3^{k+2}}0$.
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