Question

Using principle of mathematical induction prove that 6^-1 divisble by 5 .​

Answer 1

I suppose the claim is $5 \mid 6^n - 1$ for $n\in\Bbb N$.

When $n=1$, we have $6^1 - 1 = 6 - 1 = 5$, and of course 5 divides 5.

Assume the claim holds for $n=k$, that $5 \mid 6^k - 1$. We want to use this to show it holds for $n=k+1$, that $5 \mid 6^{k+1} - 1$.

We have

$6^{k+1} - 1 = \left(6^{k+1} - 6\right) + \left(6 - 1) = 6\left(6^k - 1\right) + 5$

Since $5 \mid 6^k - 1$, we can write $6^k - 1 = 5\ell$ for some integer $\ell$. Then

$6^{k+1} - 1 = 6\cdot5\ell + 5 = 5(6\ell + 1)$

which is clearly divisible by 5. QED