Editing Repunit (section)

== Properties ==
* Any repunit in any base having a composite number of digits is necessarily composite.  For example,
*: ''R''<sub>35</sub><sup>(''b'')</sup> = {{loop|35|1}} = {{loop|5|1}} × 1{{loop|6|{{loop|4|0}}1}} = {{loop|7|1}} × 1{{loop|4|{{loop|6|0}}1}},
:since 35 = 7 × 5 = 5 × 7.  This repunit factorization does not depend on the base-''b'' in which the repunit is expressed.
:Only repunits (in any base) having a prime number of digits can be prime. This is a [[Necessity and sufficiency|necessary but not sufficient condition]].  For example,
:: ''R''<sub>11</sub><sup>(2)</sup> = 2<sup>11</sup> − 1 = 2047 = 23 × 89.
* If ''p'' is an odd prime, then every prime ''q'' that divides ''R''<sub>''p''</sub><sup>(''b'')</sup> must be either 1 plus a multiple of 2''p,'' or a factor of ''b'' − 1. For example, a prime factor of ''R''<sub>29</sub> is 62003 = 1 + 2·29·1069. The reason is that the prime ''p'' is the smallest exponent greater than 1 such that ''q'' divides ''b<sup>p</sup>'' − 1, because ''p'' is prime. Therefore, unless ''q'' divides ''b'' − 1, ''p'' divides the [[Carmichael function]] of ''q'', which is even and equal to ''q'' − 1. 
*Any positive multiple of the repunit ''R''<sub>''n''</sub><sup>(''b'')</sup> contains at least ''n'' nonzero digits in base-''b''.
* Any number ''x'' is a two-digit repunit in base x − 1.
* The only known numbers that are repunits with at least 3 digits in more than one base simultaneously are 31 (111 in base-5, 11111 in base-2) and 8191 (111 in base-90, 1111111111111 in base-2). The [[Goormaghtigh conjecture]] says there are only these two cases.
* Using the [[pigeon-hole principle]] it can be easily shown that for [[Coprime integers|relatively prime]] natural numbers ''n'' and ''b'', there exists a repunit in base-''b'' that is a multiple of ''n''. To see this consider repunits ''R''<sub>1</sub><sup>(''b'')</sup>,...,''R''<sub>''n''</sub><sup>(''b'')</sup>. Because there are ''n'' repunits but only ''n''−1 non-zero residues modulo ''n'' there exist two repunits ''R''<sub>''i''</sub><sup>(''b'')</sup> and ''R''<sub>''j''</sub><sup>(''b'')</sup> with 1 ≤ ''i'' < ''j'' ≤ ''n'' such that ''R''<sub>''i''</sub><sup>(''b'')</sup> and ''R''<sub>''j''</sub><sup>(''b'')</sup> have the same residue modulo ''n''. It follows that ''R''<sub>''j''</sub><sup>(''b'')</sup> − ''R''<sub>''i''</sub><sup>(''b'')</sup> has residue 0 modulo ''n'', i.e. is divisible by ''n''. Since ''R''<sub>''j''</sub><sup>(''b'')</sup> − ''R''<sub>''i''</sub><sup>(''b'')</sup> consists of ''j'' − ''i'' ones followed by ''i'' zeroes, {{nowrap|1=''R''<sub>''j''</sub><sup>(''b'')</sup> − ''R''<sub>''i''</sub><sup>(''b'')</sup> = ''R''<sub>''j''−''i''</sub><sup>(''b'')</sup> × ''b''<sup>''i''</sup>}}. Now ''n'' divides the left-hand side of this equation, so it also divides the right-hand side, but since ''n'' and ''b'' are relatively prime, ''n'' must divide ''R''<sub>''j''−''i''</sub><sup>(''b'')</sup>.
* The [[Feit–Thompson conjecture]] is that ''R''<sub>''q''</sub><sup>(''p'')</sup> never divides ''R''<sub>''p''</sub><sup>(''q'')</sup> for two distinct primes ''p'' and ''q''.
* Using the [[Euclidean Algorithm]] for repunits definition: ''R''<sub>1</sub><sup>(''b'')</sup> = 1; ''R''<sub>''n''</sub><sup>(''b'')</sup> = ''R''<sub>''n''−1</sub><sup>(''b'')</sup> × ''b'' + 1, any consecutive repunits ''R''<sub>''n''−1</sub><sup>(''b'')</sup> and ''R''<sub>''n''</sub><sup>(''b'')</sup> are relatively prime in any base-''b'' for any ''n''.
* If ''m'' and ''n'' have a common divisor ''d'', ''R''<sub>''m''</sub><sup>(''b'')</sup> and ''R''<sub>''n''</sub><sup>(''b'')</sup> have the common divisor ''R''<sub>''d''</sub><sup>(''b'')</sup> in any base-''b'' for any ''m'' and ''n''. That is, the repunits of a fixed base form a [[Divisibility sequence|strong divisibility sequence]]. As a consequence, If ''m'' and ''n'' are relatively prime, ''R''<sub>''m''</sub><sup>(''b'')</sup> and ''R''<sub>''n''</sub><sup>(''b'')</sup> are relatively prime. The Euclidean Algorithm is based on ''gcd''(''m'', ''n'') = ''gcd''(''m'' − ''n'', ''n'') for ''m'' > ''n''. Similarly, using ''R''<sub>''m''</sub><sup>(''b'')</sup> − ''R''<sub>''n''</sub><sup>(''b'')</sup> × ''b''<sup>''m''−''n''</sup> = ''R''<sub>''m''−''n''</sub><sup>(''b'')</sup>, it can be easily shown that ''gcd''(''R''<sub>''m''</sub><sup>(''b'')</sup>, ''R''<sub>''n''</sub><sup>(''b'')</sup>) = ''gcd''(''R''<sub>''m''−''n''</sub><sup>(''b'')</sup>, ''R''<sub>''n''</sub><sup>(''b'')</sup>) for ''m'' > ''n''. Therefore, if ''gcd''(''m'', ''n'') = ''d'', then ''gcd''(''R''<sub>''m''</sub><sup>(''b'')</sup>, ''R''<sub>''n''</sub><sup>(''b'')</sup>) = ''R<sub>d</sub>''<sup>(''b'')</sup>.