What primes p exist such that 1/p has a purely periodic decimal expansion with a period of 6?
(For example, $\displaystyle 1/11 = 0.09 \overline{09}$ has a period of 2.)
(Source based on a problem by: Sanderson Smith. No peeking!)
-Dan
What primes p exist such that 1/p has a purely periodic decimal expansion with a period of 6?
(For example, $\displaystyle 1/11 = 0.09 \overline{09}$ has a period of 2.)
(Source based on a problem by: Sanderson Smith. No peeking!)
-Dan
if $\dfrac 1 p$ has a repeating cycle of 6 then
$10^6 \left(\dfrac 1 p\right) - \left(\dfrac 1 p\right) = k, ~~ k \in \mathbb{N}$
$999999 = k p$
So we want the prime factors of 999999 which are 3,7,11,13,37
I suppose it's an open question as to whether 1/3 has a period of 6 or a period of 1. It does have a pattern that repeats every 6 digits.
Thinking about these repeating periods for 1/N .... It's pretty easy to show that the maximum period value for 1/N is N-1 (for example 1/7 has a period of 6, and 1/17 has a period of 16). But of course for many values of N the period is smaller than that - for example 1/11 has a period of 2 (0.90909..) and 1/13 has a period of 6. In considering 1/N for all primes up to 100 I find that the period of 1/N is equal to (N-1)/k, where k is some positive integer that divides N-1. I had not realized that the period is restricted like this. For example, given a prime such as 103 the only possible values for its period are: 1, 2, 3, 6, 17, 34, 51, and 102. As it turns out, the period of 1/103 is actually 34, so k=3. The value for k doesn't seem to follow a pattern:
Prime, Period, k
3, 1, 2
7, 6, 1
11, 2, 5
13, 6, 2
17, 16, 1
19, 18, 1
23, 22, 1
29, 28, 1
31, 15, 2
37, 3, 12
41, 5, 8
43, 21, 2
47, 46, 1
53, 13, 4
59, 58, 1
61, 60, 1
67, 33, 2
71, 35, 2
73, 8, 9
79, 13, 6
83, 41, 2
89, 44, 2
97, 96, 1
101, 4, 25
103, 34, 3
So I'm wondering - is there a way to determine the value of k and hence the period without actually having to do the long division of 1/N?
I agree with ebaines' original answer.
As romsek explained, we are looking for primes $\displaystyle p$ such that $\displaystyle p | (10^6-1)$, however we also want $\displaystyle p \not | (10^1-1), p \not | (10^2-1), p \not | (10^3-1)$.
So, as romsek found, the prime divisors of 999,999 are 3, 7, 11, 13, and 37. Then, the distinct prime divisors of 9, 99, and 999 are 3, 11, 37. Hence, the only primes for which $\displaystyle \dfrac{1}{p}$ has a strictly periodic decimal expansion with period exactly 6 would be $\displaystyle \dfrac{1}{7}\text{ and } \dfrac{1}{13}$, as ebaines found.
In general, let $\displaystyle A_k = \{p \in \Bbb{Z} \mid p\text{ is prime and }p\text{ divides }10^k-1\}$. Then the set of primes $\displaystyle p$ such that $\displaystyle \dfrac{1}{p}$ has a strictly periodic decimal expansion with period exactly $\displaystyle N$ would be:
$\displaystyle A_N \setminus \bigcup_{n\text{ divides }N, n\neq N} A_n$
In other words, we remove all of the primes whose inverses have periods that are actually less than $\displaystyle N$.