Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The number $1/6$ can be expressed with only two digits (and a repeat sign denoted as $^\overline{}$), $$ \frac{1}{6} = \,.1\overline{6}$$ Meanwhile, it takes 49 digits to express the number $1/221$, since a string of 49 digits repeats: $$\frac{1}{221} = .\overline{004524886877828054298642533936651583710407239819}$$ Yet for $1/223$, 222 digits repeat, giving a total of 224 digits needed to express the number.

If $f:\mathbb{Q}\rightarrow\mathbb{N}$ is a function that gives the smallest number of digits needed to express a rational number in decimal notation, what can we say about $f$?

For example, if we do not consider the negative sign to be a digit, then $f$ is an odd function. Other than that, is there any pattern to it at all?

share|cite|improve this question
223 is prime ;-) – dtldarek Jul 22 '13 at 12:54
@Cocopuffs thanks! I was duped by the first recurrence of "425". Corrected. – Doubt Jul 22 '13 at 13:04
Typo in "... digits to express ... 1/121, ...": it should be "... 1/221, ...". – John Bentin Jul 22 '13 at 13:27
This related link may be useful. – user64494 Jul 22 '13 at 14:25
$233$ is prime and $1/223$ may have a $222$-digit repetend, but one should not make the mistake of thinking the shortest repetend of $1/p$ is always $p-1$ when $p$ is prime. For example, the length of the shortest repenend of $1/3$ is $1$; for $1/11$ it is $2$; for $1/37$ it is $3$; for $101$ it is $4$; for $41$ it is $5$; and for $13$ it is $6$. ${}\qquad{}$ – Michael Hardy Jun 11 '14 at 18:36
up vote 14 down vote accepted

Consider the fraction $1/m$. Write $m=2^a 5^b v$ with $\gcd(v,10)=1$. Then the periodic part of $1/m$ has length $e$, where $e$ is the smallest positive number such that $v$ divides $10^e-1$. The non-repeating part has length $f=\max(a,b)$.

There are no easy formulas for either $e$ or $f$ in terms of $m$.

share|cite|improve this answer
We can, of course, say some things about $e$. For example, $e|\phi(v)$ which gives us an upper bound. – Thomas Andrews Jul 22 '13 at 13:26
@ThomasAndrews, we can use Carmichael's $\lambda$ instead of Euler's $\phi$ and usually get much better bounds, but both functions are hard to compute when $v$ is not prime. – lhf Jul 22 '13 at 14:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.