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If you have an integer of $n$ decimal digits that is odd and non-divisble by $5$ then what is the shortest repeating decimal it can have in its reciprocal? Can it be as small as $n^c$ digits for some fixed $c>0$?

In other words what is the smallest factor of the Carmichael function of an integer of magnitude $>10^n$ that can occur as a period if the integer is coprime to $10$?

I am looking for an analytic statement such as 'if you pick a random decimal of $n$ digits coprime to $10$ then with probability $p$ it has repeating inverse of length at most $n^c$'.

Theorem 1 here seems to say there are integers of $n$ decimal digits with carmichael function as small as $n^{c_2\log\log n}$ (http://www.math.drexel.edu/~eschmutz/PAPERS/lambda.pdf). Can we we find at least one coprime pair of $n$ decimal digits with carmichael function this small?

I am mostly looking for how small we can get and not how large.

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    I assume you mean what's the shortest period of its reciprocal? – 2017-01-03
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    @jschnei yeah corrected. – 2017-01-03
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    1/99...9 (with $n$ 9s) has period $n$. This is the best you can hope for. – 2017-01-03
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    @jschnei thank you but I am looking for more general bounds. something i can use probabilistically (say if you pick a random decimal of $n$ digits coprime to $10$ then with probability $p$ it has repeating inverse of length at most $m$ - something like that). – 2017-01-03

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In the worst case (as pointed out in the comments), the smallest period of the reciprocal of a number with $n$ decimal digits is $n$ (achieved for $\frac{1}{10^{n}-1}$). No number with $n$ decimal digits can have a smaller period, since the first nonzero entry in its reciprocal must occur in the $n$th position.

The question for average case is much harder, and depends on the resolution to Artin's conjecture. Artin's conjecture states that, if $a$ is not a square or $-1$, then $a$ is a primitive root modulo roughly 37% of the primes. Applying this to $a=10$, this means if you take a random $n$-digit prime number $p$, the period of its reciprocal is $p-1$ (which is on the order of $10^n$) with probability at least 37%. By the prime number theorem, the probability a random $n$-digit number is prime is at least $\Omega(1/n)$. Together, these facts imply that the average period length over all $n$-digit numbers is at least $\Omega(10^{n}/n)$, much larger than the polynomial bounds you propose above (probably with some extra work you can improve this to $\Omega(10^{n})$).

(Of course, this is all conditioned on Artin's conjecture. It's possible there's some non-conditional approach to show the same bounds, but I haven't been able to think of one yet.)

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    great answer but theorem 1 here seems to say there are integers of $n$ decimal digits with carmichael function as small as $n^{c_2\log\log n}$ (http://www.math.drexel.edu/~eschmutz/PAPERS/lambda.pdf). Can we we find at least one coprime pair of $n$ decimal digits with carmichael function this small? – 2017-01-03
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    What do you mean by find? Find efficiently? If you just want the order of 10 modulo this number to be small, why not pick (factors of) $10^{n}-1$? – 2017-01-03
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    I mean just existence is enough. Are there equal digit coprimes factors of such $10^n-1$? – 2017-01-03
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    well, that paper shows existence? you might want to also look at https://en.wikipedia.org/wiki/Carmichael_function#Small_values. (also, this is a stronger statement than just wanting the period of the reciprocal to be small) – 2017-01-03
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    I agree that this is stronger. There is no simple way to do other wise. Do you see a better way than bounding Carmichael? – 2017-01-03
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    What do you want to do? – 2017-01-03
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    @jschneiI I want the period to be as as small $n^c$ and gcd of the repeating portion( consider as integers) to as small as $n^{c'}$ (note $n^c$ is number of digits but not $n^{c'}$). So the $10^n-1$ trick does not work. – 2017-01-03