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We know (Period of the decimal expression for the rational number $\frac{1}{n}$ is at most $n-1$) that for an integer $n\geq 2$, the period length of the decimal expression for the rational number $\frac{1}{n}$ is at most $n-1$.

I am asking if there is a lower bound if $n>m$ where $m$ is given integer.

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For $n$ relatively prime to $10$, $\dfrac1n$ repeats after $m$ digits where $n\mid10^m-1$.

Then $n<10^m$, so the period length of the decimal expression for $\dfrac1n$ is at least $ \lceil\log_{10}n\rceil$.

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  • $\begingroup$ I wondering if there a simple proof for this or at least a link to a proof. $\endgroup$
    – Safwane
    Dec 26, 2019 at 14:40
  • $\begingroup$ think of calculating $\dfrac1n$ by long division; when the remainder is $1$ (i.e., $n\mid10^m-1$), then the process repeats itself $\endgroup$
    – J. W. Tanner
    Dec 26, 2019 at 14:49
  • $\begingroup$ Ok. Thank you very much. $\endgroup$
    – Safwane
    Dec 26, 2019 at 14:51
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Consider that $\frac 1{100}=0.01$, which has period 0 or 1 depending on your definition of the period.

In other words, the lower bound of the period length is always 0 or 1.

Or if you want a non-zero repetition, we can always find a number with period 1. We can pick for instance something like $0.001111... = \frac 19\cdot \frac 1{100}=\frac 1{900}$.

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