<s>[INST] On the period of the decimal representation of $n$ when $\gcd(n, 10) \neq 1$ Suppose that $n = 2^a5^bm$, where $n > m > 1$ are integers, with $\gcd(m, 10) = 1$, and $a, b$ are non-negative integers.
How does one show that the lengths of the periods of the decimal expansions of $1/n$ and $1/m$ are the same?

I know that if the length of the period of the decimal expansion of $m$ is $r > 0$, then $10^r\equiv 1\;(\!\!\!\!\mod m)\;$, and furthermore, that $r$ is the smallest positive integer for which this congruence holds...
 [/INST] [Originally, some of the content below was included as a sort of appendix to the original question, but once I managed to complete the proof (I think), I decided to post it as an answer.]
If the length of the period of the decimal expansion of $m$ is $r > 0$, then $10^r\equiv 1\;(\!\!\!\!\mod m)\;$, and furthermore, that $r$ is the smallest positive integer for which this congruence holds.
Therefore, for some positive integer $s$, we have $s m = 10^r - 1$.  Since $m > 1$, we have that $0 < s < 10^r$, so the decimal representation of $s$ contains at most $r$ digits, and
$$\frac{1}{m} = \frac{s}{10^r} \sum_{k=0}^\infty 10^{-kr} = \frac{s\sigma}{10^r},$$
...where, for the last expression, I've defined $\sigma = \sum_{k=0}^\infty 10^{-kr}$.
Now, define $t = 2^{\max(0,b-a)}5^{\max(0,a-b)}$, and $u = \max(a, b)$.  Then we can write
$$\frac{1}{n} = \frac{1}{2^a5^bm} = \frac{t}{10^u}\cdot\frac{1}{m} = \frac{ts\sigma}{10^{r + u}}$$
Of course, the claim is trivially true if $a = b$, so let's assume that $a \neq b$.  Hence $t > 1$ and $u > 0$.
Now, using the standard identity
$$
\sigma = \sum_{k=0}^\infty 10^{-kr} = \frac{1}{1-10^{-r}} = \frac{10^r}{10^r - 1},
$$
we define $v = 10^r - 1 = 10^r/\sigma$, and let $w$ and $x$ be such that $0 \leq x < v$ and
$$
ts = vw + x.
$$
Then,
$$
\frac{1}{n} = \frac{ts\sigma}{10^{r+u}} = \frac{vw\sigma}{10^{r+u}} + \frac{x\sigma}{10^{r+u}}
$$
But, by definition of $v$, $(v \sigma)/10^r= 1$.  Hence
$$
\frac{1}{n} = \frac{w}{10^u} + \frac{x}{10^{r+u}} \sum_{k=0}^\infty 10^{-kr}
=\frac{1}{10^u} \left( w + \frac{x}{10^r} \sum_{k=0}^\infty 10^{-kr} \right).
$$
All that remains is to show that $x > 0$, or, IOW, that $v$ does not divide $ts$.  Since $t$ is either a power of $2$ or a power of $5$, whereas $v = 10^r - 1$, it follows that $\gcd(t, v) = 1$.  Therefore, $v|ts$ only if $v|s$.  But $v = sm$, and $m > 1$, so $0< s < v$, and therefore $v\nmid s$.  So we can conclude that $v\nmid ts$, and therefore, $x > 0$.
</s>