1

I was thinking today that if some fraction $1/n$ where $n$ is an integer has a digital period of $n-1$ then it must be a cyclic number. But Wikipedia says that this does hold but only states it true for when $n$ is prime. Why question is, why does $n$ need to be prime?

Anonmath101
  • 1,818
  • 2
  • 15
  • 30
  • 1
    The length of the period of $1/n$ is a divisor of $\phi(n)$. If the length is $n-1$, then $\phi(n)=n-1$ and so $n$ is prime. – lhf Feb 04 '22 at 17:01
  • But say we did not know this, can it be said that if the period is $n-1$ then it is a cyclic repeating decimal? And what you have added just further shows that this means that it can only happen when $n$ is in fact prime. – Anonmath101 Feb 04 '22 at 17:03
  • "But say we did not know this" ... Since "this" is a mathematical theorem, if we don't know it already then it is still possible to prove it. Other than the fact that the answer-in-a-comment could have been written in the answer box (@lhf), I don't see what part of the question is unanswered. – David K Feb 04 '22 at 17:40
  • I mean that I didn’t know that the length of the period of $1/n$ was a divisor of $\phi (n) $ but yet it still seemed like what I said should be true, without knowing that $n$ is prime or not. – Anonmath101 Feb 04 '22 at 17:45
  • Since you think a number "should" be cyclic if you find that the period is $n-1,$ an obvious question to ask would be, how can you prove it? I'm surprised you didn't ask that, but perhaps you got distracted by the idea that there ought to be composite values of $n$ that satisfy the statement. But you know now that there are none. Now what? Do you want a proof that the number is cyclic? (I don't have a proof, but someone else might.) – David K Feb 05 '22 at 02:05
  • Yeah I did manage to prove it. – Anonmath101 Feb 10 '22 at 23:04

0 Answers0