Let $p$ and $q$ be integers, and $n$ be the number of digits of $q$.
In the post Does $p/q$ has at most $n-1$ zeros after a non zero number in its decimal expansion, Hagen von Eitzen shows that $\frac{p}{q}$ has no more than $n-1$ digits following a non-zero digit in its decimal expansion.
This is certainly true; yet, I am trying to see if if a tightening of this result is possible.
Consider, say, the reciprocal of the prime 347:
$\frac{1}{347}$ $=.\underline{002}881844380403458213256484149855907780979827089337175792507204610951\underline{008}6455331412103746397 \\6945244956772334293948126801152737752161383285302593659942363112391930835734870317\underline{002}881844380 \\4034582132564841498559077809798270893371757925072046109510086455331412103746397694524495677233\\ 4293948126801152737752161383285302593659942363112391930835734870317002882...$
We see that exactly $n - 1 = 2$ consecutive zeros appears twice in the repetend. (I have underlined the first three occurrences of two consecutive zeros.)
Furthermore, for every case (that I can recall checking) where $q$ contains two distinct odd prime factors, the reciprocal of $q$ contains exactly $n-1$ consecutive leading zeros after the decimal point and then strictly less than $n-1$ consecutive zeros throughout the repetend.
For instance,
$\frac{1}{1241} = \frac{1}{17 \cdot 73} = \overline{0008058017727639}$
My question is: Does anyone know if this is true in general, say for two (or more) distinct odd primes ($\ne$ 5)---and if so, how may we prove it?
Or, have I not tested enough cases?
Thank you.
You can be assured that every $q$ whose first digit is $1$ satisfies your conjecture (if I understand it correctly) because the only way for the decimal expansion to have $n-1$ consecutive zeros is if the underlying division algorithm is dividing $1$ by $q$, which only will occur at the start of the repetend of $1/q$ (in this sense, $q=1241$ is definitely not illustrative of a general pattern).
This is not so when the first digit of $q$ (call it $q_n$) is larger than $1$. This is because the repetend will have $n-1$ consecutive zeros whenever the underlying division algorithm is dividing some single digit $1, 2, \ldots, q_n$ by $q$ and if it's not $1$, then it won't lie at the start of the repetend.
For instance, $$\frac{1}{91} = 0.\overline{010989}$$