Question

Find all primes \(p\) such that the decimal expansion of \(\dfrac{1}{p}\) has a repeating pattern length of \(8\)

Answer 1

what do u think

Answer 2

I feel like we solved this the other day but I feel bad that I can't remember exactly how we ended up solving it... Something to do with the geometric series though hmm...

Answer 3

I remember we solved a similar problem for length of \(5\)... but same idea yes :)

Answer 4

Can we use WA to factor large numbers or is that not allowed haha.

Answer 5

\(73\) is one.

Answer 6

11 has the period of 2 ,check it again

Answer 7

Looks like the possible numbers are 73 and 137 and I checked them all for repeating since if p=3 doesn't count since .33333333 is really period 1 not 8, but I guess it depends on how you wanna look at it... I don't want to spoil the fun though, if @mathmath333 wants to figure it out so I won't put the really interesting part of how I got these numbers. =P

Answer 8

it is kinda amazing those are the only possible candidates for a length of 8 out of all the infinitely many primes !

Answer 9

Yeah I just realized also 1/101 has a period of 4 so that also technically doesn't count. Should be fixed now lol

Answer 10

@ganeshie8 do these numbers have a name? This is really fascinating, I think I'd like to think and play with these a lot more since it's sort of seems sort of like solving a quadratic where you have two answers.

Answer 11

What I want to know is:

1) For every period, what are all the primes

2) What are these for all bases, not just base 10

Answer 12

It seems the # of primes that share the same period is pretty random as it depends on the number of prime factors of \(10^{\text{period}}-1\)... Also we saw earlier many of these prime factors have lesser periods as well...

im reading this
http://primes.utm.edu/glossary/xpage/UniquePrime.html

Answer 13

so there are only two primes \(73\) and \(137\)

which satisfy \(10^{8} -1\pmod {prime}=0\)

Answer 14

Fascinating.

Answer 15

discrete logarithm problem ?

Answer 16

Although probably unrelated I just noticed that the two numbers with period 8 are 73 and 137 and when you add them up you get 210 which is 8 primorial, which is like a factorial but all the primes less than it, \[\Large 8 \# = 7*5*3*2=210\] Probably just a weird coincidence.

Answer 17

i see u keep making weird relations

Answer 18

hahaha

Answer 19

Unfortunately it doesn't seem to work, but there still might be some relation.

For period 6 there are two numbers, 7 and 13 and so 6#=30 but 7+13=20 so ten off. Shoot, wouldn't that be fascinating if there was a connection? The only reason I think there might be some chance is because this is all in base 10, so we have some sort of relationship going on.

Answer 20

@mathmath333 I almost forgot, did you want to see the solution?

Answer 21

yes

Answer 22

Since it repeats every 8 digits, if you'll allow me to put a letter to represent digits, we have \[\frac{1}{p} = 0.abcdefghabcdefgh...\] and really what we have is that there's this possibly eight digit number (for instance, 00000001 is acceptable since i has period 8) \[abcdefgh = n\] So if we sort of plug it into the formula below we really have: \[\frac{1}{p} = n*10^{-8}+n*10^{-16}+ \cdots\] which is really just a geometric series in 10^-8, \[\frac{1}{p} = n * \frac{10^{-8}}{1-10^{-8}} = \frac{n}{10^8-1}=\frac{n}{99999999}\] So if we rearrange this to solve for the prime we have: \[\Large p = \frac{99999999}{n}\] So from there you can just factor 10^8-1 into prime factors using wolfram alpha or try to do something fancy, but that's really all there's left to do.

Answer 23

nice work