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anticorncob28
Post  Post subject: Math question  |  Posted: Sat May 13, 2017 2:49 am
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I already know the answer to this; I want to see if you guys can do it.

Let n be an integer greater than 10 such that the decimal expansion of 1/n repeats with period (n - 1). Prove that the decimal expansion of 1/n must contain an 8.

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Olinguito
Post  Post subject: Re: Math question  |  Posted: Sun May 14, 2017 7:52 pm
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I can't prove it but one thing I can be sure of is that n must be a prime number. This is because the length of the period of recurrence in the decimal expansion of $\frac1n$ must be a divisor of $\phi(n)$ (where $\phi$ is the Euler totient function, denoting the number of positive integers not greater than than n which are relatively prime to n). This can be shown as a consequence of Euler's theorem (a generalization of Fermat's little theorem). For n > 2 (and if it's greater than 10 it is certainly greater than 2) $\phi(n)$ is not divisible by $n-1$ unless $n=p$, a prime. (Note that this is only a necessary condition, not a sufficient one: there are some primes p (e.g. p = 11) for which $\frac1p$ has a decimal-expansion period less than $p-1$.)

As for the problem, it appears that there is nothing special about the digit 8: all the digits 0–9 appear in the decimal expansion, in some order or other. I don't know why this is so but I suspect it has to do with the fact that when n is prime, the ring of integers modulo n is a field. (And the fact that for some primes p the period of recurrence is less than $p-1$ may be due to the fact that 10 is not a primitive root modulo p.) Or maybe there is a much simpler, down-to-earth explanation. :roll:

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anticorncob28
Post  Post subject: Re: Math question  |  Posted: Sun May 14, 2017 9:51 pm
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Olinguito wrote:
I can't prove it but one thing I can be sure of is that n must be a prime number. This is because the length of the period of recurrence in the decimal expansion of $\frac1n$ must be a divisor of $\phi(n)$ (where $\phi$ is the Euler totient function, denoting the number of positive integers not greater than than n which are relatively prime to n). This can be shown as a consequence of Euler's theorem (a generalization of Fermat's little theorem). For n > 2 (and if it's greater than 10 it is certainly greater than 2) $\phi(n)$ is not divisible by $n-1$ unless $n=p$, a prime. (Note that this is only a necessary condition, not a sufficient one: there are some primes p (e.g. p = 11) for which $\frac1p$ has a decimal-expansion period less than $p-1$.)

As for the problem, it appears that there is nothing special about the digit 8: all the digits 0–9 appear in the decimal expansion, in some order or other. I don't know why this is so but I suspect it has to do with the fact that when n is prime, the ring of integers modulo n is a field. (And the fact that for some primes p the period of recurrence is less than $p-1$ may be due to the fact that 10 is not a primitive root modulo p.) Or maybe there is a much simpler, down-to-earth explanation. :roll:

You're (somewhat) on the right track. There is nothing special about the digit 8, and the same general reasoning that proves it with 8 will also prove it for the digits 0 - 9.
The solution (or at least my solution) does not involve fields, or any abstract algebra at all. It can be understood with high school algebra and knowing modular arithmetic.

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anticorncob28
Post  Post subject: Re: Math question  |  Posted: Fri May 19, 2017 3:00 am
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Okay, here's a hint:
What happens if you move the decimal point?
If nobody answers then I'll post the answer, and maybe add further math challenges on this thread.

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