Show that $10^{n(p-1)}\equiv 1\pmod{\! 9p}$ for $p\ge 7$

Question

I need to prove that for each prime $p \ge 7$ and for each $n \in\Bbb N$

$$10^{n(p-1)} \equiv 1 \pmod {9p}$$

What I've tried:

I know $10$ is coprime to $9$ and $p$, so it is coprime to $9p$.

I know that $\,\varphi (9p) = \varphi (9)\cdot \varphi (p) = 6(p-1)$

and now I'm stuck. I guess I'm missing some basic rule.

Any help or hint will be appreciated.

Answer 1

By Fermat's little theorem, since $(10,p)=1$,

$$10^{n(p-1)}\equiv \left(10^{p-1}\right)^n\equiv 1^n\equiv 1\pmod{\! p}$$

Also $$10^{n(p-1)}\equiv 1^{n(p-1)}\equiv 1\pmod{\! 9}$$

Since $(9,p)=1$, we have $$9,p\mid 10^{n(p-1)}-1\iff 9p\mid 10^{n(p-1)}-1$$

Answer 2

Hint $\ \ \begin{align} 10^{\,\color{#c00}e}\equiv 1\!\!\!\pmod{q}\\10^f\equiv 1\!\!\!\pmod{p}\end{align}\ \Rightarrow\ 10^{\,\color{#0a0}{{\rm lcm}(e,f)}}\!\equiv 1\pmod{\!{\rm lcm}(p,q)}$

Yours is: $\ \color{#c00}{e}=1,\ $ therefore $\ \color{#0a0}{{\rm lcm}(e,f)} = f$