Patterns in Repeating Decimals

Date: 08/06/2003 at 19:45:49
From: Patrick J. Igoe
Subject: Patterns in repeating decimals

1/7 = .142857... 2/7 = .285714... 3/7 = .428571... 4/7 = .571428...
5/7 = .714285... 6/7= .857142...

For each fraction, the repeating decimal is the same basic sequence 
of six digits [142857], which starts at slightly different starting 
points [for 1/7, "1" is starting point ... and for 6/7, "8" is 
starting point] and wraps around.  Also, the starting point does not 
simply move successively down the line of the same basic sequence 
[from 1 to 4, from 4 to 2, from 2 to 8]. Rather, the starting point 
always moves to the next highest/largest available number within the 
same basic sequence [1 to 2, 2 to 4, 4 to 5, 5to 7, 7 to 8]. Why is this?

Other fractions are slightly different.  For #/13, there appear to 
two (2) different six-digit sequences.
 1/13 = .076923...      2/13 = .153846...
 3/13 = .230769...      5/13 = .384615...
 4/13 = .307692...      6/13 = .461538...
 9/13 = .692307...      7/13 = .538461...
10/13 = .769230...      8/13 = .615384...
12/13 = .923076...     11/13 = .846153...

Any thoughts on this? Can it be generalized?
Thanks for your help, 
Patrick

Date: 08/07/2003 at 08:34:06
From: Doctor Jacques
Subject: Re: Patterns in repeating decimals

Hi Patrick,

This is no accident.

If n is not divisible by 2 or 5, the length of the minimal period of 
the decimal expansion of 1/n is a divisor of phi(n), where phi(n) is 
Euler's function (if n is prime, phi(n) = n-1). This is a consequence 
of Euler's Theorem.

If n is prime, and the length of the period is equal to n-1 (the 
largest possible value), the fractions m/n, where m and n are 
relatively prime, will have the same period, starting at a different 
point.

This is the case for n = 7, where the expansion of 1/7 has period 6. 
Another example is n = 19:

  1/19 = 0.052631578947368421...
  2/19 = 0.105263157894736842...

where the period is equal to 18.

The length of the period is equal to the smallest exponent k > 0 such 
that 10^k = 1 (mod n).

If the period has length k < phi(n), there may be up to phi(n)/k 
different patterns. For example, as you noticed, the period of 1/13 
has length 6, and there are 12/6 = 2 different patterns.

Another example is n = 9, phi(9) = 6, and the period has length 1. In 
this case, we have 6 different periods (we only consider the 
numerators relatively prime to 9, since otherwise we are actually 
dealing with multiples of 1/3):

 1/9 = 0.111...
 2/9 = 0.222...
 4/9 = 0.444..
 5/9 = 0.555..
 7/9 = 0.777..
 8/9 = 0.888...

If the denominator is divisible by 2 and / or 5, you must first 
divide it to remove those factors in order to consider the period 
length. In that case, the periodic part will not start right after 
the decimal point.

Does this help?  Write back if you'd like to talk about this 
some more, or if you have any other questions.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 08/07/2003 at 12:55:12
From: Patrick J. Igoe
Subject: Thank you (patterns in repeating decimals)

Thank you for your answer, 
Patrick