Prove that $2^{9693}-1$ divisible by $7$ Prove that $2^{9693}-1$ divisible by $7$, by more than one way.
my try... that, the power divisible by $3$ so it's divisible by $7$ like $2^3,2^6,2^{12}$ and I think it's wrong.

Claim: For any $n\in\mathbb N$, $2^{3n}-1$ is divisible by $7$.
Proof: Induction. For $n=1$, $2^{3\times 1}-1=7$. Now suppose that the claim is true for $1,2,\ldots,n$. Observe that 
\begin{align*}
2^{3(n+1)}-1=2^{3n+3}-1=2^{3n}\times 2^3-1=8\times 2^{3n}-1=8\times 2^{3n}-8+7=8\times(2^{3n}-1)+7.
\end{align*}
By the induction hypothesis, $2^{3n}-1$ is divisible by $7$, and thus so is $8\times(2^{3n}-1)$. Divisibility by $7$ is preserved by the addition of the extra term $7$. Therefore, $2^{3(n+1)}-1$ is divisible by $7$ and the proof is complete. $\quad\blacksquare$
Now apply the claim for $n=3\mathrm,231$.
