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The number $142,857$ is widely known as a cyclic number, meaning consecutive multiples are cyclic permutations, i.e.

$1 × 142,857 = 142,857$

$2 × 142,857 = 285,714$

$3 × 142,857 = 428,571$

and so on.

142857 is the repeating unit of $\frac{1}{7} = 0.\overline{142857}$ and in fact, every prime for which $10$ is a primitive root will generate a cyclic number (if we allow $0$ as a first digit, for example $0588235294117647$ which is the repeating unit of $\frac{1}{17}$). These primes are called full reptend primes.

From what I've read, it seems that there is a bijection between full reptend primes and cyclic numbers: a number is cyclic if and only if it is the repeating unit for the reciprocal of a full reptend prime.

I was able to find a proof for the if part. I was wondering if anyone can provide a proof for the only if part.

Edit: It's been a while since I posed this question and I still haven't found a proof. I've added a bounty in hopes of prompting some interest.

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    http://math.stackexchange.com/questions/443/why-is-the-decimal-representation-of-1-7-cyclical2011-08-12
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    @DJC That question is concerned with the cyclic nature of the numbers whereas I'm more interested in a proof that every cyclic number is the reciprocal of a full reptend prime.2011-08-12
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    If $n$ is the cyclic number and $d$ the number of digits (counting the $0$ if present), then we need to prove $(10^d-1)/n$ is a whole prime and also a primitive root modulo $10$. No doubt both parts will involve exploiting $n$'s cyclical nature and the arithmetic properties of its cycles.2011-08-12
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    @EuYu : I think this would be inventing a new proof for the statement.2012-01-31
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    @Iyengar Meaning that the statement is currently unproven?2012-01-31

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