# 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.

• Welcome to math SE. Have a look at mathjax for your mathematical expressions. – Alain Remillard Feb 13 '20 at 19:32
• This is a great question. – Mike Feb 13 '20 at 19:39

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 $$10$$.

• Well but why is that the rule though? – Mike Feb 13 '20 at 19:35
• It might be instructive to explain why this rule holds. – Cheerful Parsnip Feb 13 '20 at 19:36
• @Mike You will find a proof here. – José Carlos Santos Feb 13 '20 at 19:43
• The point is that by Fermat's "little" theorem, if $p$ is a prime other than $2$ or $5$ we have $10^{p-1} \equiv 1 \mod p$. – Robert Israel Feb 13 '20 at 19:51
• After reading up on Fermat's Little Principle, I now understand that $(10^(p-1)-1)/p$ will be an integer. That's great, because that integer is the exact sequence of digits that repeats, and the integer is $p-1$ digits long. F's LP doesn't predict that in the case of $1/3$, there is one repeating digit, or in the case of $1/11$, there are two. This formula gives 33 and 909090909 respectively. – kpres Feb 14 '20 at 1:03

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.

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.

• I'm sorry, I'm having trouble understanding the syntax you are using. I think you are explaining how to predict the length of the period. Based on what you wrote, I notice that when I use $L=2$ and then put it into the formula $(10^(L+1)-10)/p$, only the prime reciprocals of length 2 will be integers; that is, 3 and 11. This is a great clue. Thank you for this insight! – kpres Feb 14 '20 at 18:03