It is very easy.
1/9 = 0.1111111 ...
1/99 = 0.0101010 ...
1/999 = 0.0010010 ...
...
so,
For example, in case [076923] (length=6)
numerator = 76923
denominator = 999999 (6 digits)
computing reduction,
76923/999999 = 1/13
is the answer.
Now, multiply both sides of equation by the denominator, then,
76923 * 13 = 999999 = 9 * 111111
so, we find a non-trivial factor 13 except trivial factors, 3, 11 and 111.
The length e of repeating sequence of 1/p is given by the following formula,
e is the smallest number where 10e≡1 (mod p)
(i.e. the order of 10 under mod p.j
In case of 13,
101≡10 (mod 13)
102≡ 9 (mod 13)
103≡12 (mod 13)
104≡ 3 (mod 13)
105≡ 4 (mod 13)
106≡ 1 (mod 13)
so, e=6. At the above example,
111111 = (106-1)/9 = (10e-1)/9
so, in general,
p | (10e-1)/9
and because the length of repeating sequence is at most p-1,
e | p-1
so,
p | (10p-1-1)/9
This is equal to Fermat's theorem.
The number 11...11=(10n-1)/9 (n times iteration of "1") is called as
repunit, denoting by Rp. For example, 111111 = R6 (above case).
Especially, when the length is prime, it does not have trivial factors, so,
are the main concerns.
Harvey Dubner researched (2007) primes up to 200000 and found when
p = 2, 19, 23, 317, 1031, 49081, 86453, 109297
then repunits are also prime.
(last three cases, p=49081, 86453, 109297, then Rp are probable prime.)
And Maksym Voznyy found that R270343 is a probable prime (July 15, 2007).
More generally,
(bn-1)/(b-1)
are also under reseach. Which is called as general repunit, base-b repunit.
"10,n-" in
The Cunningham Project
is the current results of factorization of repunit.
The following are under computation.
On March 03, 2008, there are "First Five Holes".
10,241- c229
10,257- c241
10,263- c216
10,269- c233
10,271- c214
so, all the repunits under 239 were completely factorized.
For factors of repunits Rp where p is prime under 100 are here.
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