If $n$ is coprime to 10, then $1/n$ produces a repeating decimal. EDIT: I have changed the title (twice). I am getting my terms and symbols jumbled up in my head. I hope that I have asked a clear question now. 

Here are three statements that I believe to be true, but have only been able to prove them in one direction.


CASE 1: Iff $n = 2^x5^y$ then $1/n$ produces a terminating decimal.
CASE 2: Iff n is coprime to 10, 1/n produces a purely repeating decimal
CASE 3: Iff $n = 2^x5^yn'$, where $n'$ is coprime to 10, then 1/n produces a repeating decimal after a non-repeating section.


I already proved case 1 both ways, but I'm stuck on case 2. I think if I can prove case 2, then case 3 would be very easy. 

Question:
Are the above statements true? If so, what are some hints to help me solve case 2?

My work so far:

Assume that $n$ is coprime to 10. 

$\implies 10^k \bmod n = m$, where $m \in \{1, \dots, 9\}, \forall k \in \mathbb{N}$ 

$\implies$ I'm really not sure where to go next...  

I should add that I never studied number theory before. This is for fun, on my own time, so simple explanations are better for me :)
Yes, all three statements are true.

Terminating expansions: A positive real number $x$ has a terminating decimal expansion iff there exists $k$ such that $10^k x \in \mathbb N$. Now suppose $x = 1/n$. Then 
$$\begin{align*}
&amp;\exists k \in \mathbb N : 10^k \frac{1}n \in \mathbb N \\
\iff &amp;\exists k, \ell \in \mathbb N : 10^k = n \ell
\end{align*}$$
That is, $1/n$ has a terminating decimal expansion iff it divides $10^k$ for some $k$. That is, iff its only prime divisors are $2$ and $5$. This proves the first claim.

Periodic expansions. A positive real number $x&lt;1$ has a periodic decimal expansion iff there exists $k&gt;0$ such that $10^k x - x \in \mathbb N$, i.e. $(10^k-1) x \in \mathbb N$. Now take $x = 1/n$, with $n &gt; 1$. Then this is equivalent to saying that there exists $k&gt;0$ such that $n$ divides $10^k - 1$. We want to show that this is equivalent to $10$ being coprime to $n$.

$n$ coprime to $10$ $\implies$ periodic expansion: Consider the numbers $10^0, 10^1, 10^2, \ldots$ and their remainders upon division by $n$. Because there are infinitely many numbers and only finitely many possible remainders, ...


   by the pigeonhole principle, there exist $p, q \in \mathbb N$ with $p &lt; q$ and such that $10^p$ and $10^q$ have the same remainder upon division by $n$.


That is, ...


   $n$ divides $10^q - 10^p$.


We have $10^q - 10^p = 10^{p} (10^{q-p}-1)$.
Because $n$ divides this number, and $n$ is coprime to $10$, it must be that ...


   $n$ divides $10^{q-p} - 1$.


Thus $1/n$ has periodic expansion.

Periodic expansion $\implies$ $n$ coprime to $10$: By assumption, there exist $k, \ell \in \mathbb N$ with $10^k -1 = n \ell$, and $k \geq 1$. Suppose $n$ and $10$ are not coprime. Then there exists $d &gt; 1$ that divides both $10$ and $n$. But then $d$ divides ...


  $$10^k - n \cdot \ell = 1\,,$$


... a contradiction with $d &gt; 1$. Hence $n$ is coprime to $10$.

Eventually periodic expansions. The decimal expansion of any rational number is finite or eventually periodic, so it must be that the natural numbers $n$ not covered by the previous two cases are precisely those such that $1/n$ has an infinite, not immediately periodic decimal expansion.
Counterexample to the title:

Prove that, in general if $1/n$ produces a repeating decimal, then $n = 2^x5^y$.

Let $n=7$.  $$\frac{1}{7} = 0.\overline{142857}$$

The number $7$ is prime, so $7 \ne 2^x 5^y$.

UPDATE

Hey, the title changed.  This actually was a valid response at the time!