4.50. Find all natural numbers $n$ for which $2^{n}-1$ is divisible by 7.
4.50. It is clear that $2^{3}=8 \equiv 1(\bmod 7)$. Therefore, $2^{3 k} \equiv 1(\bmod 7)$, $2^{3 k+1} \equiv 2(\bmod 7)$ and $2^{3 k+2} \equiv 4(\bmod 7)$. Consequently, $2^{n}-1$ is divisible by 7 if and only if $n$ is divisible by 3.