Prove that $7^n - 1$ is divisible by 6

Question

I know that I have to prove this with the induction formula. If proved the first condition i.e. $n=6$ which is divisible by $6$. But I got stuck on how to proceed with the second condition i.e $k$.

Answer 1

Induction step:n+1.

$7^{n+1} -1 = 7\cdot 7^{n} -1=$

$ (6+1)(7^n) -1=$

$6 \cdot 7^n +(7^n -1).$

By hypothesis $(7^n-1)$ is divisible by $6$, hence the above sum is divisible by $6.$

Answer 2

We have $$7\equiv 1\mod 6$$ then $$7^n\equiv 1^n=1\equiv 1\mod 6$$ so $$7^n-1\equiv 0 \mod 6$$

Answer 3

Hint: $7^{n+1}-1=7^{n+1}-7^n+7^n-1=6\times 7^n+7^n-1$.

Answer 4

Let suppose that the result is true for $n=k$ i.e $7^k-1$ is divisible by $6$.

Then: $$7^{k+1}-1=7 \times (7^k-1)+7-1=7 \times (7^k-1)+6$$ but $(7^k-1)$ is divisible by $6$ so $7 \times (7^k-1)$ is also divisible by $6$ and finally $(7^k-1)+6$ is divisible by $6$.

Answer 5

We have: $$7^n-1=(7-1)(7^{n-1}+7^{n-2}+\dots +7+1) = 6(7^{n-1}+7^{n-2}+\dots +7+1) $$

Which definitely is divisible by $6$.