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