In mathematics, the transposable integers are integers that permute or shift cyclically when they are multiplied by another integer n {\displaystyle n} . Examples are:
142857 × 3 = 428571 (shifts cyclically one place left) 142857 × 5 = 714285 (shifts cyclically one place right) 128205 × 4 = 512820 (shifts cyclically one place right) 076923 × 9 = 692307 (shifts cyclically two places left) These transposable integers can be but are not always cyclic numbers. The characterization of such numbers can be done using repeating decimals (and thus the related fractions), or directly.
General For any integer coprime to 10, its reciprocal is a repeating decimal without any non-recurring digits. E.g. 1⁄143 = 0.006993006993006993... While the expression of a single series with vinculum on top is adequate, the intention of the above expression is to show that the six cyclic permutations of 006993 can be obtained from this repeating decimal if we select six consecutive digits from the repeating decimal starting from different digits. This illustrates that cyclic permutations are somehow related to repeating decimals and the corresponding fractions. The greatest common divisor (gcd) between any cyclic permutation of an m-digit integer and 10m − 1 is constant. Expressed as a formula,
gcd ( N , 10 m − 1 ) = gcd ( N c , 10 m − 1 ) , {\displaystyle \gcd \left(N,10^{m}-1\right)=\gcd \left(N_{c},10^{m}-1\right),}
where N is an m-digit integer; and Nc is any cyclic permutation of N. For example,
gcd(091575, 999999) = gcd(32×52×11×37, 33×7×11×13×37) = 3663 = gcd(915750, 999999) = gcd(157509, 999999) = gcd(575091, 999999) = gcd(750915, 999999) = gcd(509157, 999999)
If N is an m-digit integer, the number Nc, obtained by shifting N to the left cyclically, can be obtained from:
N c = 10 N − d ( 10 m − 1 ) , {\displaystyle N_{c}=10N-d\left(10^{m}-1\right),\,}
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 10m−1. For examples the related fractions to the above cyclic permutations are thus:
091575⁄999999, 915750⁄999999, 157509⁄999999, 575091⁄999999, 750915⁄999999, and 509157⁄999999. Reduced to their lowest terms using the common gcd, they are:
25⁄273, 250⁄273, 43⁄273, 157⁄273, 205⁄273, and 139⁄273. That is, these fractions when expressed in lowest terms, have the same denominator. This is true for cyclic permutations of any integer.
Fraction method
Integral multiplier An integral multiplier refers to the multiplier n being an integer:
An integer X shift right cyclically by k positions when it is multiplied by an integer n. X is then the repeating digits of 1⁄F, whereby F is F0 = n 10k − 1 (F0 is coprime to 10), or a factor of F0; excluding any values of F which are not more than n. An integer X shift left cyclically by k positions when it is multiplied by an integer n. X is then the repeating digits of 1⁄F, whereby F is F0 = 10k - n, or a factor of F0; excluding any values of F which are not more than n and which are not coprime to 10. It is necessary for F to be coprime to 10 in order that 1⁄F is a repeating decimal without any preceding non-repeating digits (see multiple sections of Repeating decimal). If there are digits not in a period, then there is no corresponding solution. For these two cases, multiples of X, i.e. (j X) are also solutions provided that the integer i satisfies the condition n j⁄F < 1. Most often it is convenient to choose the smallest F that fits the above. The solutions can be expressed by the formula:
X = j 10 p − 1 F {\displaystyle X=j{\frac {10^{p}-1}{F}}}
where p is a period length of 1⁄F; and F is a factor of F0 coprime to 10. E.g, F0 = 1260 = 22 × 32 × 5 × 7. The factors excluding 2 and 5 recompose to F = 32 × 7 = 63. Alternatively, strike off all the ending zeros from 1260 to become 126, then divide it by 2 (or 5) iteratively until the quotient is no more divisible by 2 (or 5). The result is also F = 63. To exclude integers that begin with zeros from the solutions, select an integer j such that j⁄F > 1⁄10, i.e. j > F⁄10. There is no solution when n > F.
Fractional multiplier An integer X shift left cyclically by k positions when it is multiplied by a fraction n⁄s. X is then the repeating digits of s⁄F, whereby F is F0 = s 10k - n, or a factor of F0; and F must be coprime to 10. For this third case, multiples of X, i.e. (j X) are again solutions but the condition to be satisfied for integer j is that n j⁄F < 1. Again it is convenient to choose the smallest F that fits the above. The solutions can be expressed by the formula:
X = j s 10 p − 1 F {\displaystyle X=js{\frac {10^{p}-1}{F}}}
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