For prime $p>5$ and positive integer $k<p$, show that the decimal expansion of $k/p$ consists of $p-1$ repeating decimal digits

Question

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 $2$.

Answer 1

If $10^{p-1} = 1 + mp$, where $1 < m < 10^{p-1}$, that says $$ \frac{1}{p} = \frac{m}{10^{p-1}-1} = \frac{m}{10^{p-1}(1-10^{-p+1})} = \sum_{j=1}^\infty \frac{m}{10^{j(p-1)}} $$ The decimal representation of this sum consists of "$0.$" followed by the digits of $m$ (padded at the front with zeros if necessary to length $p-1$) repeated: each term of the sum represents one block of $p-1$ digits.

Nobody said that $p-1$ has to be the smallest period. You can consider $4/11$ as $0.(36)(36)(36)\ldots$ with period $2$, but you could also write it as $0.(3636363636)(3636363636)\ldots$ with period $10=11-1$. An example where $p-1$ is the smallest period is $$1/7 = 0.(142857)(142857)\ldots$$