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Consider, for example, $\frac{1}{N} = \frac{1}{195} = \frac{1}{3\cdot5\cdot13} = 0.0051282051282051282... = 0.0\overline{051282}$,

compared to, say,

$\frac{1}{N} = \frac{1}{77} = \frac{1}{7\cdot11} = 0.012987012987... = 0.\overline{012987}$.

QUESTION: Under what conditions will the repetend of $1/N$ include all of the zeros following the decimal point, such as in the example $\frac{1}{77}$? Is there a particular class of integers $N$ which this must always happen? I have noticed that this seems to be the case when $N$ is a semiprime, but it is not the case, as we see above, when $N$ has three prime factors.

Bill Dubuque
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DDS
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1 Answers1

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Let's say that $1/N=0.00\dots 0\overline{a_1a_2\dots a_k}$, where there are $m$ zeroes, and $A$ is the number represented by $a_1a_2\dots a_k$ in base $10$. Then $$0.\overline{a_1a_2\dots a_k}=\frac{A}{10^k-1}$$ and so $$\frac 1N=\frac{A}{10^m(10^k-1)}.$$ As a result, $N$ should divide $10^m(10^k-1)$. So, $m$ can't be $0$ if $2$ or $5$ divide $n$; see if you can convince yourself that $m$ can be $0$ in all other cases.

Carl Schildkraut
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  • Thank you for your answer. I can now see that $m$ can be $0$ in all other cases; however, it seems to me that if neither 2 nor 5 divides $N$, then $m$ MUST be $0$. Am I correct in this conclusion? Thank you again. – DDS Sep 20 '20 at 04:20
  • @samuelbowditch It depends how you define repeating. For example, you're perfectly allowed to say that $\frac1{11}=0.0\overline{90}$ instead of $0.\overline{09}$, but this is clearly not the most economical representation -- such a representation would be unique, and will necessarily have $m=0$. – Carl Schildkraut Sep 20 '20 at 04:32