This can't be right. Factorisation is, legendarily, a hard problem, unsolved (in realistic time frames) for large input. But more than that, it's harder than division is.

Finding the length of the nonrepeating and repeating parts of a rational number's decimal expansion can be done just by doing the division. That would appear to be evidence that it is not harder than division?

Edit: from the wikipedia section you link:

> If k = 2^a·5^b·n where n > 1 and n is not divisible by 2 or 5, then the length of the transient of 1/k is max(a,b) and the period [of the repetend, presumably] is r, where r is the smallest integer such that 10^r ≡ 1 (mod n)

All integers (except those whose only prime factors are 2 and 5) satisfy this condition, so this would appear to be exactly the method desired? Repeatedly dividing by 5 is much, much easier than factoring is. Taking the discrete log of 1... could be equivalent to factoring.

From your edit I guess that you already realized this, but the hard part here is that for 1 / q you would have to perform up to q divisions of integers of size n = log₂(q) which of course takes time O(2ⁿ) even assuming the divisions are O(1).

I still remember that quite well because that was probably the first and only time that I was unable to find someone stating that we do not know any better, and not only the Wikipedia article but everywhere I looked there was no such statement to be found. I eventually gave up after a couple of evenings but it obviously kept stuck somewhere in my mind and that is why I wrote the initial comment.

Finding the order of an element in a group is a hard problem on par with factoring http://mathworld.wolfram.com/GroupOrder.html

I'm not sure about cases where the state space doesn't form a multiplicative group, but I doubt that it gets much easier.