Came across a funny little problem last week. Explain the following:
For primes [tex]p \ge 7[/tex] , the decimal expansion of [tex]\frac{1}{p}[/tex] is non-terminating. Is there a formula for the number of repeating digits?
[tex]\frac{1}{7} = 0.\overline{142857}[/tex] 6 digits
[tex]\frac{1}{11} = 0.\overline{09}[/tex] 2 digits
[tex]\frac{1}{13} = 0.\overline{076923}[/tex] 6 digits
[tex]\frac{1}{17} = 0.\overline{0588235294117647}[/tex] 16 digits
and my favourite twin primes
[tex]\frac{1}{41} = 0.\overline{02439}[/tex] 5 digits
[tex]\frac{1}{43} = 0.\overline{023255813953488372093}[/tex] 21 digits
Yes there is – Theorem 135 from Hardy & Wright (4th edition) looks at the decimal for any rational. In the case of 1/p it says that:
a) there are no non-recurring digits
b) there are v recurring digits where v is the order of 10 mod p ie the smallest solution of 10^v=1 mod p
Yes, that’s the answer but I didn’t know that it appeared in Hardy & Wright. Thanks for the reference. I really have to get a copy of that book. Although, I remember vaguely that someone is working on a new version of that classic, so I’m waiting.