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

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 $$2$$.

• Maybe $p-1$ does not need to be the minimal period, so since $2\mid10$ that is sufficient. – YiFan Oct 17 '19 at 23:29
• Consider the consequences of $10^{p-1}\equiv1\pmod{p}$ and how one computes $1/p$ in base ten. (Look at the remainders.) – robjohn Oct 17 '19 at 23:35
• @robjohn but still why (p-1) in the example? It seems not right based on the example I gave. – Jerry_Ge Oct 17 '19 at 23:38
• @Jerry_Ge: Did you read YiFan's comment? $0.\overline{6363636363}$ has a period of $10$, but also a period of $2$. – robjohn Oct 18 '19 at 1:51
• This might help: math.stackexchange.com/q/3295196/597411 – C. Melton Oct 18 '19 at 3:44

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 $$1/7 = 0.(142857)(142857)\ldots$$

• Thanks for the clarification! 1. Curious how you get from m/(10^(p-1)(1-10^{-p+1}) to the summation using geometric series? [Seems not intuitive for me] 2. why p <= 5 will not work? – Jerry_Ge Oct 18 '19 at 16:13
• @Jerry_Ge: $p=3$ works. But $p=2$ and $p=5$ don't work, because the decimal expansions of $k/2$ and $k/5$ terminate. – TonyK Nov 23 '19 at 19:44