Trying to prove:
Let $p > 5$ be a prime and let $k$ be any positive integer $< p$. Show that the decimal expansion of $\frac{k}{p}$ consists of (p-1) repeating decimal digits. (hint: use Fermat's Little Theorem and Geometric Series).
I was trying to understand the theorem using an example but I am not very sure about why (p-1) repeating decimal digits.
E.g suppose $p = 11$, k = $4$, $5$, $7$.
- $\frac{4}{11} = 0.36363636363636\ldots$
- $\frac{5}{11} = 0.454545454545454545\ldots$
- $\frac{7}{11} = 0.63636363636363636\ldots$
So where does $p-1 = 10$ come from? It seems the repeating length is always $2$.