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Is there a way to calculate the number of non-repeating digits that precede the periodic repeating portion of a decimal expansion? For example:

1/6 = 0.1666.... (there is 1 non repeating digit) **(Correction) 1/12 = 0.08333... (there are 2 non repeating digits) 7/12 = 0.58333....(there are 2 non repeating digits) 1/96 = 0.01041666..(there are 5 non repeating digits)

Do any forumulas exist for predicting the maximum length n, of the number of non repeating digits preceding the repeating portion?

I know that if the denominator of a fraction is n, the maximum length of the repeating periodic portion is n-1. Must also the length of the preceding portion before the cycle be n-1?

Thank you!

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  • $\begingroup$ 1/6 is 0.1666... ; 1/7 is 0.142857... $\endgroup$ Commented Oct 11, 2010 at 1:53
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    $\begingroup$ Otherwise, this might be more suitable for math.stackexchange.com $\endgroup$ Commented Oct 11, 2010 at 1:54
  • $\begingroup$ 1/6 = 0.16666...thanks for the correction $\endgroup$
    – user9934
    Commented Oct 11, 2010 at 1:59
  • $\begingroup$ That 1/11 is really 1/12. $\endgroup$ Commented Oct 11, 2010 at 2:28

2 Answers 2

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When one writes an irreducible fraction $m/n$ as a periodic digit number all one does is to write

$m/n=\frac{a}{999...9000.00}$

So the number of digits before the period is the maximum of the power of $2$ and $5$ in $n$, i.e. wirting $n=2^\alpha 5^\beta k$ with $k$ relatively prime to $10$, the number of digits before the period is $\max\{\alpha, \beta \}$.

I think that this will follow for free from the following lemma, whose proof is trivial:

Lemma: if gcd$(k,10) =1$ then $k$ has a multiple of the form $999...9$.

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  • $\begingroup$ So stated another way, the number of non-repeating decimals is equal to the number of times the denominator is divisible by either 5 or 2. For example, 1/12, 12 is divisible by 2, 2 times, and there are 2 non-repeating digits. 1/96, 96 is divisible by 2 - 5 times... Do you recommend a good number theory book that explains this in more depth? Thank you Nick! $\endgroup$
    – user9934
    Commented Oct 11, 2010 at 3:28
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Suppose we deal with base $B\in\mathbb{Z}$. The trick is to consider the distinct primes that divide $B$. If $a/b$ is in lowest terms, then the prime factorization of $b$ can be sorted into primes that divide $B$ and primes that don't. If $b=uv$, where every prime dividing $u$ divides $B$ and no prime dividing $v$ divides $B$, then $u\mid B^n$ for some integer $n$, and $(v, B)=1$. Then in base $B$, the expansion of $a/b$ satisfies the following: (a) The smallest integer $n$ such that $u\mid B^n$ is the number of non-repeating digits that precede the periodic repeating portion; (b) The order of $B$ (mod $v$) - it exists since $(v,B)=1$ and theoretically divides $\phi(v)$ - is the length of the periodic repeating portion.

The special case where $B=10$ is what is usually used. In that case, the distinct primes dividing $B$ are $2$ and $5$. Thus $u=2^\alpha 5^\beta$ and the smallest $n$ such that $u\mid 10^n$ would be $\max\{\alpha,\beta\}$. So Nick S was right.

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