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Thread: Calculating the period of a decimal

  1. #1
    Super Member Showcase_22's Avatar
    Sep 2006
    The raggedy edge.

    Calculating the period of a decimal

    Find the delay and period of \frac{60}{336}.

    How I started:

    \frac{60}{336}=\frac{5}{2^2 \cdot 7}.

    Hence the delay \mu=\max \{0,2 \}=2.

    I've run into a bit of a problem when I try and find the delay.

    I need to find n s.t 10^n=1 \bmod 7.

    Comparing this to Euler's theorem \Rightarrow n= \phi(7)=6\Rightarrow \nu=6..

    My mark scheme also agrees with this result. However, can it always be done in this way? ie. will the delay always be reduced to 10^n=1 \bmod p where p is prime?
    Last edited by Showcase_22; Jun 3rd 2009 at 05:00 AM.
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  2. #2
    Super Member malaygoel's Avatar
    May 2006
    you are talking about period..right.
    Please check your is confusing!!!!
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  3. #3
    Member alunw's Avatar
    May 2009
    The answer to your second question is not quite. To find the repeating decimal for 1/q where q has no factors of 2 or 5 you need to find 10^n mod q = 1 and the answer will be the lowest common multiple of the n's that solve 10^n mod p^a = 1 where p is prime and a is the highest power of the prime that divides q.
    For example 1/27 has period 3 even though 1/3 has period 1: you really have to solve 10^n mod 27 =1
    1/91 has period 6 because both 1/7 and 1/13 have period 6.
    But 1/119 has period 48 because 119=17*7 and 17 has period 16 and 7 has period 6

    Euler's totient function has nothing to do with the period of a decimal, so the fact that 7 has period 6 and phi(7)=6 must be a coincidence. The definition of the totient function has nothing to do with our use of 10 as our number base. 1/7 has period 1 in base 7, period 3 in base 4.
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