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In base 10, the recurring bits of the fractions $\frac{1}7,\ldots,\frac{6}7$ are cyclic permutations of each other. e.g. $$\frac{1}{7}=0.(142857)$$

$$\frac{2}{7}=0.(285714)$$

$$\frac{3}{7}=0.(428571)$$

In which bases does there exist $n$ such that the recurring bits of the fractions $\frac{1}{n},\ldots,\frac{n-1}{n}$ are cyclic permutations of each other?

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Cyclic numbers exist for a base $b$ iff there is a prime $p$ such that $b$ is a primitive root mod $p$. Artin's conjecture says that there are plenty of examples. However, there are no cyclic numbers for bases that are perfect squares. See http://en.wikipedia.org/wiki/Cyclic_number.

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    can you please provide a proof or a link to the proof of "Cyclic numbers exist for a base b iff there is a prime p such that b is a primitive root mod p" ? thanks in advance.2012-01-17
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    @NikhilBellarykar, the expansion of each $k/p$ should start at a different position in the expansion of $1/p$. This implies that the expansion of $1/p$ has a full-length period that is, $p-1$, which can only happen if $b$ is a primitive root mod $p$. If $n$ is not a prime, then $1/n$ cannot have a full-length period. See http://en.wikipedia.org/wiki/Repeating_decimal#Other_properties_of_repetend_lengths.2012-01-17
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    Thanks. But 13 has primitive roots and it does not have only one set of cyclic numbers but two. So are you talking about only the existence of any number of cyclic numbers or only one of them here? I guess some condition for existence of unique cyclic number can also be helpful, what say?2012-01-17
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    @NikhilBellarykar, 10 is not a primitive root modulo 13, because $13\mid 10^6-1$. That's why the length of the recurring period of decimals in $k/13$ has length six. As opposed to twelve that you would get, if ten were primitive.2012-01-17
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    @NikhilBellarykar, not sure what you mean. But for 13, take say 6 or 7, which are primitive roots mod 13. Then in base 6 or base 7, 1/13 is cyclic.2012-01-17
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    @lhf: ah I was bit confused, now it's clear somewhat. What I meant was, for 7, we have only one number whose permutations are the decimal expansions of all fractions from $1/7$ to $6/7$, whereas, for 13, we have 2 such numbers. What I would like to know is that whether any condition exists that tells us how many cyclic numbers are there for a given base, assuming they exist?2012-01-17
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    @Jyrki Lahonten: Got your point. thanks for clarification.2012-01-17