The reciprocal of many prime numbers p in base 10 have a set of repeating digits p-1. Why? I've noticed that the reciprocal of many prime numbers have a curious number of digits. For example. 1/7 has 6 repeating digits, 1/17 has 16 repeating digits, and 1/47 has 46. There's a rule here that makes this pattern, and I haven't quite figured it out. Does anyone have some insight on this?
There are a few other numbers that follow the pattern subversively. For example, 1/3 has two repeating digits: 33. 1/11 has ten repeating digits: 0909090909. The number 1/13 has a set of six that repeats twice, which makes a set that's 12 digits long.
For reference, here is a website that lists reciprocals of numbers 2 through 70, including non-primes.
https://thestarman.pcministry.com/math/rec/RepeatDec.htm
The ones that break the rule are 1/2 and 1/5, but they seem to be a matched pair, where the 2 and 5 switch places in the equation 1/X = Y/10
What is the rule that makes this pattern? I'm very curious.
 A: Actually, in the case of $\frac13$, you have one repeating digit: $3$.
And the rule is: if $p$ is a prime number other than $2$ and $5$, then the period of the digital expansion of $\frac1p$ divides $p-1$. For instance, when $p=13$, the period is $6$ (as you wrote) and $6\mid 12$. But it could not be, say, $5$ or $8$.
And why $2$ and $5$ are exceptions? Because we work in base $10$ and $2$ and $5$ are the prime factors of $10$.
A: This is a consequence of Fermat's little theorem,
according to which $p$ divides $10^{p-1}-1$ if $p$ is not a factor of $10$ (i.e., $2$ or $5$). 
If you picture the long division for $\dfrac 1p$, since the remainder after division of $10^{p-1}$ by $p$ is $1$, 
the cycle of digits in the quotient repeats after ${p-1}$ digits.  
A: Let $q$ be a prime s.t. $(q,10)=1$ and let $\ell$ be a period of $\frac{10}{q}$ ; i.e., the $j$-th and the $j+\ell$-th digit of $\frac{10}{q}$ are the same for all positive integers $j$. Then observe that $q|(10^{\ell+1}-10)$, which implies that $10^{\ell} \equiv_q 1$. Can you finish from here.
