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It is easy to show that $0.1\overline{04}$ (repedent : $04$) is $\dfrac{104-1}{990} = \dfrac{103}{990}$.

But it is quiet difficult for me to show that 'if the denominator of an irreducible fraction is $990$, the length of repedent must be 2 when this fraction is changed to repeating decimal' is true or not.

Interestingly, I think this question is related to number theory although it is about to rational number.

I'd like to know a proof if it is true or some couterexamples if it is false.

Bill Dubuque
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Paris
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  • It just comes down to decimals of the form $0.\overline{ab}=\frac{ab}{99}$, so to push the repetition back, you have to multiply the denominator by $10$ – Rushabh Mehta Mar 02 '20 at 20:12
  • The condition is the denominator of 'Irreducible fraction' is 990. – Paris Mar 02 '20 at 20:20
  • In general, if the denominator of an irreducible fraction is $\ 2^a×5^b×c\ $ with $\ \gcd(2,c)=\gcd(5,c)=1\ $ and $\ c\ge3\ ,$ then the length of the repedent is the smallest positive integer $\ r\ $ such that $\ 10^r−1\ $ is divisible by $\ c\ .$ For a denominator of $\ 990=2×5×99\ ,$ $\ 10^2−1\ $ is divisible by $\ 99\ $ but $\ 10^1−1\ $ isn't, so this criterion tells you that the length of the repedent is $2.$ – lonza leggiera Apr 06 '25 at 11:13
  • Reversing dupe's proof here writes any rational as a repeating decimal. Here we get $$\begin{align} \frac{n}{990} &=, \frac{1}{10}\frac{n}{99},,\ \ n = 99q+\color{#c00}r,\ \rm by\ dividing,\ n\div 99\[.4em] &=, \frac{1}{10}\left[q + \frac{\color{#c00}r}{99}\right],\ \ \color{#c00}{0\le r < 99}\[.4em] &=, 0.01, [q + 0.\overline{01}, \color{#c00}r,]\[.4em] &=, 0.01, [q + 0,.:!\overline{\color{#c00}r},]\ \end{align}\qquad$$

    so once we get past the digits from $,q,,$ it repeats the two-digits in $,\color{#c00}{r}., \ $

     – Bill Dubuque Apr 06 '25 at 17:26
  • Note that if the two digits in $,\color{#c00}r,$ are equal then cycle has period-length $1$ (not $2).\ \ $ – Bill Dubuque Apr 06 '25 at 17:30

1 Answers1

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$\dfrac{1}{990} = .0\overline{01}$

If you multiply this by any whole number from $1-99$, it will repeat that whole number over two digits. For example,

$\dfrac{57}{990} = .0\overline{57}$

$\dfrac{99}{990}$ Does not repeat.

$\dfrac{99}{990} = .1$

No matter what fraction you have, you can 'break off' as many $\dfrac{99}{990}$ as needed to make it $\dfrac{99n}{990} + \dfrac{a}{990}$ where $a$ will be less than $99$. For example,

$\dfrac{577}{990} = \dfrac{5\times99}{990} + \dfrac{82}{990} = 5\times.1 + .0828282...$

Since the part broken off will always be a non-repeating multiple of $.1$, the repeating part will always be 2 digits.

Thirdy Yabata
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user78090
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