I discovered that the decimal representation of 15/17 consists of an endlessly repeating sequence of 16 digits. I wonder what rational number has the longest sequence of repeating digits?

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- Jul 22nd 2010, 01:30 PMStevenBrownRepeating decimals
I discovered that the decimal representation of 15/17 consists of an endlessly repeating sequence of 16 digits. I wonder what rational number has the longest sequence of repeating digits?

- Jul 22nd 2010, 01:36 PMStevenBrown
Thinking more about this, it occurs to me that a rational number expressed as a fraction having arbitrarily long numerator and denominator that differ by 1 can have an astronomically long sequence of repeating digits in the decimal representation.

- Jul 22nd 2010, 03:04 PMwonderboy1953
- Jul 22nd 2010, 03:14 PMundefined
You might find useful this reference (also the other subheadings)

Wikipedia - Repeating decimal - subheading Other properties of repetend lengths

You might also like to think backwards; given any positive decimal expansion with a repetend, we can find unique positive integers p,q such that gcd(p,q)=1 and p/q has the given value. - Jul 22nd 2010, 05:03 PMStevenBrown
It occurred to me that a repeating decimal such as 15/17 =

0.8823529411764705 8823529411764705 8823529411764705 ...

is "going fractal," in the sense that the same pattern repeats endlessly on smaller and smaller scales. - Jul 22nd 2010, 08:21 PMSoroban
Hello, StevenBrown!

You discovered this on your own? . . . Good for you!

Keep exploring . . . there's some fascinating stuff out there!

Quote:

I discovered that the decimal representation of $\displaystyle \frac{15}{17}$

consists of an endlessly repeating sequence of 16 digits.

I wonder what rational number has the longest sequence of repeating digits?

Considerable exploration has already been done on this subject.

Here are some basics that I remember . . .

If $\displaystyle p$ is a prime, the decimal for $\displaystyle \frac{1}{p}$ has a $\displaystyle p-1$ digit repeating cycle.

However, some primes can have even shorter cycles.

For example: .$\displaystyle \begin{Bmatrix}\frac{1}{11}\text{ has a 2-digit cycle:} &0.\overline{09} \\ \\[-3mm]

\frac{1}{13}\text{ has a 6-digit cycle:} & 0.\overline{076923} \\ \\[-3mm]

\frac{1}{37} \text{ has a 3-digit cycle:} & 0.\overline{027}\end{Bmatrix}$

What determines the length $\displaystyle n$ of the cycle for prime reciprocals?

It is the least $\displaystyle n$ for which $\displaystyle 10^n-1$ is divisible by $\displaystyle p.$

In baby-talk, consider a string of 9's.

Begin dividing by a prime $\displaystyle p.$

When it "comes out even", stop.

The number of 9's is the length of the cycle.

Example: $\displaystyle \frac{1}{7}$

The first time the division stops is: .$\displaystyle 999999 \div 7 \,=\,142857$

. . Therefore, $\displaystyle \frac{1}{7}$ has a 6-digit cycle.

Example: $\displaystyle \frac{1}{37}$

The first time the division stops is: .$\displaystyle 999 \div 37 \,=\,27$

. . Therefore, $\displaystyle \frac{1}{37}$ has a 3-digit cycle.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here's another interesting feature . . .

. . $\displaystyle \begin{array}{ccc}

\frac{1}{7} &=& 0.\overline{142857} \\ \\[-3mm]

\frac{2}{7} &=& 0.\overline{285714} \\ \\[-3mm]

\frac{3}{7} &=& 0.\overline{428571} \\ \\[-3mm]

\frac{4}{7} &=& 0.\overline{571428} \\ \\[-3mm]

\frac{5}{7} &=& 0.\overline{714285} \\ \\[-3mm]

\frac{6}{7} &=& 0.\overline{857142} \end{array}$

The last five decimals are cyclic arrangements of the first one.