a time to repeat

rob mahurin

2025-09-30 22:25

Consider the following decimal expansions:

$$\begin{aligned} \textstyle \frac{1}{ 1 } &= 1.0 & \textstyle \frac{1}{ 11 } &= 0.[09] & % 0.09090909090909091 \\ \textstyle \frac{1}{ 21 } &= 0.[04761\,9] \\ % 0.047619047619047616 \\ % \textstyle \frac{1}{ 2 } &= 0.5 & \textstyle \frac{1}{ 12 } &= 0.08[3] & % 0.08333333333333333 \\ \textstyle \frac{1}{ 22 } &= 0.0[45] \\ % 0.045454545454545456 \\ % \textstyle \frac{1}{ 3 } &= 0.[3] & % 0.3333333333333333 \\ \textstyle \frac{1}{ 13 } &= 0.[07692\,3] & % 0.07692307692307693 \\ \textstyle \frac{1}{ 23 } &= 0.[04347\,82608\,69565\,21739\,13] \\ % 0.043478260869565216 \\ % \textstyle \frac{1}{ 4 } &= 0.25 & \textstyle \frac{1}{ 14 } &= 0.0[71428\,5] & % 0.07142857142857142 \\ \textstyle \frac{1}{ 24 } &= 0.041[6] \\ % 0.041666666666666664 \\ % \textstyle \frac{1}{ 5 } &= 0.2 & \textstyle \frac{1}{ 15 } &= 0.0[6] & % 0.06666666666666667 \\ \textstyle \frac{1}{ 25 } &= 0.04 \\ % \textstyle \frac{1}{ 6 } &= 0.1[6] & % 0.16666666666666666 \\ \textstyle \frac{1}{ 16 } &= 0.0625 & \textstyle \frac{1}{ 26 } &= 0.0[38461\,5] \\ % 0.038461538461538464 \\ % \textstyle \frac{1}{ 7 } &= 0.[14285\,7] & % 0.14285714285714285 \\ \textstyle \frac{1}{ 17 } &= 0.[05882\,35294\,11764\,7] & % 0.058823529411764705 \\ \textstyle \frac{1}{ 27 } &= 0.[037] \\ % 0.037037037037037035 \\ % \textstyle \frac{1}{ 8 } &= 0.125 & \textstyle \frac{1}{ 18 } &= 0.0[5] & % 0.05555555555555555 \\ \textstyle \frac{1}{ 28 } &= 0.035[71428\,5] \\ % 0.03571428571428571 \\ % \textstyle \frac{1}{ 9 } &= 0.[1] & % 0.1111111111111111 \\ \textstyle \frac{1}{ 19 } &= 0.[05263\,15789\,47368\,421] & % 0.05263157894736842 \\ \textstyle \frac{1}{ 29 } &= 0.[03448\,27586\,20689\,65517\,24137\,931] \\ % 0.034482758620689655 \\ % \textstyle \frac{1}{ 10 } &= 0.1 & \textstyle \frac{1}{ 20 } &= 0.05 & \textstyle \frac{1}{ 30 } &= 0.0[3] \\ % 0.03333333333333333 \\ \end{aligned}$$

Each of these either terminates, like $\frac18 = 0.125$, or repeats, like $\frac{1}{27} = 0.[037]$. What determines the length of these repeating sequences?

Rational numbers must terminate or repeat

Decimal expansions are produced by long division, which is a process of repeatedly taking quotients and remainders. An example:

Here we have

Twenty-seven goes into one zero times, with remainder one.
Multiply the remainder by ten. Twenty-seven goes into ten zero times, with remainder ten.
Multiply the remainder by ten. Twenty-seven goes into one hundred three times, with remainder nineteen.
Multiply the remainder by ten. There are seven twenty-sevens in one hundred ninety, with remainder one.
Multiply the remainder by ten. We've already asked whether twenty-seven goes into ten, so we've entered a loop.

An integer $b$ can in principle have $b-1$ different nonzero remainders, so $a/b$ can have at most $b-1$ repeating digits in its decimal representation. In the table above, we see that $1/7$, $1/17$, $1/19$, $1/23$, and $1/29$ have this "maximum-length" representation. But other representations, such as $1/11$ and $1/27$, are substantially shorter.

Of course, if we ever hit a remainder of zero, then the decimal expansion terminates.

Terminating decimals must have denominators $2^j 5^k$

If a rational number $a/b$ has a decimal representation which terminates, then we can say

$$ \frac ab = \frac c {10^n} $$

for integers $a,b,c$. There must exist some integer $d$ so that $ad = c$ and $bd = 10^n$. Because the product $bd$ is a factor of ten, the integers $b$ and $d$ can have only two and five as prime factors.

Immediately-repeating decimals can be written with an "all nines" denominator

Let's limit ourselves to repeating decimals which start immediately at the decimal point, like $1/7$. These can be written as

$$\begin{aligned} \frac ab &= \frac{c}{10^n} + \frac{c}{10^{2n}} + \frac{c}{10^{3n}} + \cdots \end{aligned}$$

for integers $a,b,c$, where the $n$ repeating digits are the digits of $c$. We can rewrite as a geometric series,

$$\begin{aligned} \frac ab &= c\times\left( \frac{1}{10^n} + \frac{1}{10^{2n}} + \frac{1}{10^{3n}} + \cdots \right) \\ &= c\sum_{i=1}^{\infty}\left(\frac{1}{10^n}\right)^i = \frac{c}{10^n - 1} \end{aligned}$$

This suggests, for instance, that $142\,857 \times 7 = 999\,999$, because $142\,857$ are the digits in the decimal expansion of $1/7$. You can even imagine using this as a way to factorize numbers of the "all-nines" form $10^n-1$, by asking what numbers have decimal expansions of length $n$.

Non-immediately-repeating decimals are annoying

Expansions like $\frac{1}{15} = 0.0[6]$, which eventually repeat but have some cruft at the front, are annoying.

I think these always have a factor of two or five in the denominator, and that fractions without any factors of two or five in the denominator always have representations that repeat immediately. I haven't tried to prove this.

You can always make the repeat start at the decimal point by multiplying by enough powers of ten. If there are $\ell$ leading digits that don't repeat, and then the repeating pattern has period $p$, then we can write our number as

$$\begin{aligned} \frac ab &= \frac c {10^\ell} + \frac{1}{10^\ell}\cdot\frac{d}{10^p - 1} \end{aligned}$$

for integers $a,b,c,d$. Then

$$\begin{aligned} \frac ab \times 10^\ell &= c + \frac{d}{10^p-1} = \frac ef \end{aligned}$$

starts repeating at the decimal point. Note that no all-nines number $10^p-1$ is ever divisible by any factor of two or five, so the new denominator $f$ doesn't have two or five as a factor.

Is that a proof? It doesn't feel like one.

Repeating digit patterns do cyclic group-y things

Consider now the fractions with denominator seven:

$$\begin{aligned} \textstyle \frac{ 1 }{ 7 } &= { 142\,857 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 2 }{ 7 } &= { 285\,714 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 3 }{ 7 } &= { 428\,571 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 4 }{ 7 } &= { 571\,428 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 5 }{ 7 } &= { 714\,285 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 6 }{ 7 } &= { 857\,142 } / \textcolor{lightgray}{ 999\,999 } \\ \end{aligned}$$

Our list of digits $142\,857$ has the maximum allowed length, which means we must reach it in every fraction with denominator seven. For the fractions with denominator thirteen, there are two digit lists of length six:

$$\begin{aligned} \textstyle \frac{ 1 }{ 13 } &= \textcolor{red}{ \ 76\,923 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 2 }{ 13 } &= \textcolor{blue}{ 153\,846 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 3 }{ 13 } &= \textcolor{red}{ 230\,769 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 4 }{ 13 } &= \textcolor{red}{ 307\,692 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 5 }{ 13 } &= \textcolor{blue}{ 384\,615 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 6 }{ 13 } &= \textcolor{blue}{ 461\,538 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 7 }{ 13 } &= \textcolor{blue}{ 538\,461 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 8 }{ 13 } &= \textcolor{blue}{ 615\,384 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 9 }{ 13 } &= \textcolor{red}{ 692\,307 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 10 }{ 13 } &= \textcolor{red}{ 769\,230 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 11 }{ 13 } &= \textcolor{blue}{ 846\,153 } / \textcolor{lightgray}{ 999\,999 } \\ \textstyle \frac{ 12 }{ 13 } &= \textcolor{red}{ 923\,076 } / \textcolor{lightgray}{ 999\,999 } \\ \end{aligned}$$

With denominator eleven, the ten repeating numerators are given by five distinct digit lists with length two:

$$\begin{aligned} \textstyle \frac{ 1 }{11} &= 0.[09] & \textstyle \frac{ 2 }{11} &= 0.[18] & \textstyle \frac{ 3 }{11} &= 0.[27] & \textstyle \frac{ 4 }{11} &= 0.[36] & \textstyle \frac{ 5 }{11} &= 0.[45] \\ \textstyle \frac{ 10}{11} &= 0.[90] & \textstyle \frac{ 9 }{11} &= 0.[81] & \textstyle \frac{ 8 }{11} &= 0.[72] & \textstyle \frac{ 7 }{11} &= 0.[63] & \textstyle \frac{ 6 }{11} &= 0.[54] \\ \end{aligned}$$

The fractions with denominator twenty-one are interesting. There are twenty numerators, repeating with length six. Eight of those numerators we have seen already: the six with form $3n/21$, and the two with form $7n/21$. The remaining twelve are divided into two groups with length six:

$$\begin{aligned} \textstyle \frac{ 1 }{ 21 } &= 0.[\textcolor{red}{ 047619 }] \\ \textstyle \frac{ 2 }{ 21 } &= 0.[\textcolor{blue}{ 095238 }] \\ \textstyle \frac17 = \frac{ 3 }{ 21 } &= 0.[\textcolor{green}{ 142857 }] \\ \textstyle \frac{ 4 }{ 21 } &= 0.[\textcolor{red}{ 190476 }] \\ \textstyle \frac{ 5 }{ 21 } &= 0.[\textcolor{blue}{ 238095 }] \\ \textstyle \frac27 = \frac{ 6 }{ 21 } &= 0.[\textcolor{green}{ 285714 }] \\ \textstyle 1/3 = \frac{ 7 }{ 21 } &= 0.[\textcolor{lightblue}{ 333333 }] \\ \textstyle \frac{ 8 }{ 21 } &= 0.[\textcolor{blue}{ 380952 }] \\ \textstyle \frac37 = \frac{ 9 }{ 21 } &= 0.[\textcolor{green}{ 428571 }] \\ \textstyle \frac{ 10 }{ 21 } &= 0.[\textcolor{red}{ 476190 }] \\ \textstyle \frac{ 11 }{ 21 } &= 0.[\textcolor{blue}{ 523809 }] \\ \textstyle \frac47 = \frac{ 12 }{ 21 } &= 0.[\textcolor{green}{ 571428 }] \\ \textstyle \frac{ 13 }{ 21 } &= 0.[\textcolor{red}{ 619047 }] \\ \textstyle 2/3 = \frac{ 14 }{ 21 } &= 0.[\textcolor{lightblue}{ 666666 }] \\ \textstyle \frac57 = \frac{ 15 }{ 21 } &= 0.[\textcolor{green}{ 714285 }] \\ \textstyle \frac{ 16 }{ 21 } &= 0.[\textcolor{red}{ 761904 }] \\ \textstyle \frac{ 17 }{ 21 } &= 0.[\textcolor{blue}{ 809523 }] \\ \textstyle \frac67 = \frac{ 18 }{ 21 } &= 0.[\textcolor{green}{ 857142 }] \\ \textstyle \frac{ 19 }{ 21 } &= 0.[\textcolor{red}{ 904761 }] \\ \textstyle \frac{ 20 }{ 21 } &= 0.[\textcolor{blue}{ 952380 }] \\ \textstyle \frac{ 21 }{ 21 } &= 0.[\textcolor{lightgray}{ 999999 }] \\ \end{aligned}$$

It's cute how these have to all fit together, without any leftovers. The prime-checking for the all-nines denominators has the smell of Fermat's Little Theorem about it. Why do expansions with period three and five not show up until after a bunch of expansions with period six?

And how do these things act when you look at "decimal" expansions in other bases? We can measure these periods without actually doing expansions:

base = 10
for p in range(0,100):
    for n in range(1,p):
        q,r = divmod(base**n-1,p)
        if r == 0:
            print(f'in base {base}, 1/{p} has period {n}');
            break

There are also some funny things happening with powers. In decimal, $1/27$ has period three, and $1/81$ has period nine. One way to think of that would be that each new power of three in the denominator triples the length of the sequence? It's likewise the case that $1/49$ has period $6\times 7 = 42$. But I'm kind of horsing around with that stuff.