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where b is the number base (10 for decimal), and p is a prime that does not divide b. (Primes p that give cyclic numbers in base b are called full reptend primes or long primes in base b). For example, the case b = 10, p = 7 gives the cyclic number 142857, and the case b = 12, p = 5 gives the cyclic number 2497.
Therefore, the base b expansion of / repeats the digits of the corresponding cyclic number infinitely, as does that of / with rotation of the digits for any a between 1 and p − 1. The cyclic number corresponding to prime p will possess p − 1 digits if and only if p is a full reptend prime.
A cyclic number [1] [2] is a natural number n such that n and φ(n) are coprime. Here φ is Euler's totient function. An equivalent definition is that a number n is cyclic if and only if any group of order n is cyclic. [3] Any prime number is clearly cyclic. All cyclic numbers are square-free. [4] Let n = p 1 p 2 …
[1] [2] For example, 1193 is a circular prime, since 1931, 9311 and 3119 all are also prime. [3] A circular prime with at least two digits can only consist of combinations of the digits 1, 3, 7 or 9, because having 0, 2, 4, 6 or 8 as the last digit makes the number divisible by 2, and having 0 or 5 as the last digit makes it divisible by 5. [4]
where d is the first digit of N and m is the number of digits. This explains the above common gcd and the phenomenon is true in any base if 10 is replaced by b, the base. The cyclic permutations are thus related to repeating decimals, the corresponding fractions, and divisors of 10 m −1. For examples the related fractions to the above cyclic ...
The smallest base greater than binary such that no three-digit narcissistic number exists. 80: Octogesimal: Used as a sub-base in Supyire. 85: Ascii85 encoding. This is the minimum number of characters needed to encode a 32 bit number into 5 printable characters in a process similar to MIME-64 encoding, since 85 5 is only slightly bigger than 2 ...
If p is a proper prime ending in a 1, that is, if the repetend of 1 / p is a cyclic number of length p − 1 and p = 10h + 1 for some h, then each digit 0, 1, ..., 9 appears in the repetend exactly h = p − 1 / 10 times. For some other properties of repetends, see also. [15]
If you square the last three digits and subtract the square of the first three digits, you also get back a cyclic permutation of the number. [citation needed] 857 2 = 734449 142 2 = 20164 734449 − 20164 = 714285. It is the repeating part in the decimal expansion of the rational number 1 / 7 = 0. 142857.