Re: [math-fun] 1/7 etc.

On Mar 17, 2018, at 3:13 PM, Simon Plouffe <simon.plouffe@gmail.com> wrote:

we can remark that 142857 is symmetrical, 142 + 857 = 999

so one just has to remember half of the period in order to know all of the digits, 1/13 = 076923 , 076+923 = 999 too.

nice - since 7 and 13 are both divisors of 1001.

this phenomena is true whenever p <> 2 or p <> 11 and the period is even.

wait - why does that imply that p divides 10^k+1 for some k? that’s the same as saying that 10 mod p has even order in the multiplicative group, right? Huh... - Cris