Divisibility of number $s\perp 10$

Question

I found a property that for $s\perp 10$ we can say that there exists such $k$ that $$ 10^k \equiv 1 \pmod s $$ but I cannot prove it. Can you help me in this task.

My approach:

When $s\perp 10$ then we know that there exists such $x,y$ that $10x+sy=1$. So we can state that $10x\equiv1 \;\;(mod\;s)$. But I don't know what do next.

Answer 1

Let $a,b$ be naturals with $a\perp b$. Consider the infinite sequence $$a\bmod b,\ a^2\bmod b,\ a^3\bmod b,\ \ldots$$ As only $b$ distinct values are possible, some repetition must occur, i.e., we can find naturals $1\le i<j$ such that $a^i\equiv a^j\pmod b$. Then with $k:=j-i>0$, we have $a^i\equiv a^ia^k\pmod b$. As $a\perp b$, this implies $1\equiv a^k\pmod b$.

Answer 2

By the Euler’s theorem, such $k$ exists and $k=\varphi(s)$.