Let $k$ be a strictly positive integer. Find all strictly positive integers $n$ such that $3^{k}$ divides $2^{n}-1$.
Let $k \geq 1$ be an integer such that $3^{k}$ divides $2^{n}-1$. By reasoning modulo 3, we see that $n$ is even. Let us write $n=2 m$ with $m>0$. Then $3^{k}$ divides $4^{n}-1$. Since 3 divides $4-1$, we can apply the LTE theorem:

$$
v_{3}(4-1)+v_{3}(n)=v_{3}\left(4^{n}-1\right) \geq k
$$

We deduce that $v_{3}(n) \geq k-1$. Thus $2 \times 3^{k-1}$ divides $n$.

Conversely, the same reasoning gives us that $3^{k}$ divides $2^{n}-1$ if $2 \times 3^{k-1}$ divides $n$.