Prove 11 does not divide $3^{3k-1}+5*3^k$ for any odd k. First I did an induction proof that it does work for even k. Then I started the proof as so.
Suppose there exists a k of the form 2n+1, s.t 11 divides $3^{3k-1}+5*3^k$.
After some algebra I can arrive at this point $5*2^{6n-1}+3(2^{6n-1}+5*3^{2n})$
Since I proved separately this works for even k, the right side sum if of that form, $(2^{6n-1}+5*3^{2n})$ is divisible by 11. So I believe if I can somehow prove that 11 does not divide any power of 2, I would have finished the proof. However I don't know how to do that.
Someone may have to fix the tags as I'm not entirely sure what is appropriate here, sorry.
 A: HINT
If $11|2^m$ for $m\in\mathbb{N}$ it would mean that $11|2$ which is a contradiction. 
This follows from the fact that if a prime $p$ divides the product of $2$ numbers, say $p|ab$ then $p|a$ or $p|b$. Apply this to $a=2^{m-1}, b=2$.
So if all else is correct your proof is finished.
A: A general proof for $k$ even and odd.$$3^{3k-1}+5*3^k=3^k(3^{2k-1}+5)$$ for which if $11$ divides the expression then $11$ should divide $3^{2k-1}+5$. This is not possible because the subgroup generated by $3$ is equal to $\{3,9,5,4,1\}$ and the five possible sums give, respectively the classes $$8,3,10,9,6$$ any of them is equal to zero modulo $11$.
A: $3^{3k-1}+5\times 3^k=3^{k-1}(3^{2k}+15)=3^{k-1}(x^2+15)$ with $x=3^k$
Also $x^2+15\equiv x^2+4\pmod{11}$
$\begin{array}{l}
k=0: & 3^0\equiv 1\pmod{11} & x^2+4\equiv 5\pmod{11}\\
k=1: & 3^1\equiv 3\pmod{11} & x^2+4\equiv 13\equiv 2\pmod{11}\\
k=2: & 3^2\equiv 9\pmod{11} & x^2+4\equiv 85\equiv 8\pmod{11}\\
k=3: & 3^3\equiv 5\pmod{11} & x^2+4\equiv 29\equiv 7\pmod{11}\\
k=4: & 3^3\equiv 4\pmod{11} & x^2+4\equiv 20\equiv 9\pmod{11}\\
k=5: & 3^5\equiv 1\pmod{11} & \text{and it cycles there...}\\
\end{array}$
So $x^2+4$ is never a multiple of $11$ for any $x=3^k$, and since $3^{k-1}$ is neither divisible by $11$ we have our result for any $k\ge 1$.
