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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...

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I think you are overcomplicating. The factors of $2$ and $5$ create a non-repeating initial segment, and the repeating bit is all down to the factors which are co-prime to $10$. Hardy and Wright has a chapter which deals with this and other things like "normal" numbers –  Mark Bennet Jun 29 '13 at 19:13
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As Mark Bennet said, multiplying a number by 2 or by 5 just doesn't change its period in base 10. –  chubakueno Jun 29 '13 at 19:22
    
@chubakueno: your assertion is what I'm trying to prove... –  kjo Jun 29 '13 at 19:50

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up vote 0 down vote accepted

[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, $x > 0$.

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