Let m be a positive integer relatively prime to 10. Then the period length of 1/m is the multiplicative order of 10 modulo m -- i.e. we look in the group (Z/mZ)*. Note that if p is an odd prime, that even though the multiplicative,a =order of 10 mod p might be small, the multiplicative order of 10 mod p^2 is overwhelmingly likely to be p*a (except when we have something like a Wieferich prime). If m,n are relatively prime, then by the Chinese remainder theorem the multiplicative order of 10 mod mn is the lcm of the multiplicative orders of 10 mod each of m and n. Victor On Mon, Sep 29, 2014 at 5:53 PM, James Propp <jamespropp@gmail.com> wrote:
I had assumed that if you multiply an eventually repeating decimal of period m and an eventually repeating decimal of period n, you get an eventually repeating decimal whose period is bounded by some polynomial function of m and n. But today I learned from Henry Cohn that that's not true: the period length can be exponentially large in m and n. More specifically, 1/(10^n-1) repeats with period n, but its square repeats with period 10^n-1 or thereabouts.
I'm sure someone has written about the repeat-length for 1/(10^n-1)^2; can anyone provide a reference?
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