Transposable integer

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 10^m − 1 is constant. Expressed as a formula,

${\displaystyle \gcd \left(N,10^{m}-1\right)=\gcd \left(N_{c},10^{m}-1\right),}$

where N is an m-digit integer; and N_c 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 N_c, obtained by shifting N to the left cyclically, can be obtained from:

${\displaystyle N_{c}=10N-d\left(10^{m}-1\right),\,}$

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 10^m−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.

The direct algebra approach to the above cases integral multiplier lead to the following formula:

1. ${\displaystyle X=D{\frac {10^{m}-1}{n10^{k}-1}},}$

   where m is the number of digits of X, and D, the k-digit number shifted from the low end of X to the high end of n X, satisfies D < 10^k.

   If the numbers are not to have leading zeros, then n 10^k − 1 ≤ D.

2. ${\displaystyle X=D{\frac {10^{m}-1}{10^{k}-n}},}$

   where m is the number of digits of X, and D, the k-digit number shifted from the high end of X to the low end of n X, satisfies:

   1. ${\displaystyle D<{\frac {10^{k}}{n}}-1,}$
   2. and the 10-part (the product of the terms corresponding to the primes 2 and 5 of the factorization) of 10^k − n divides D.

      The 10-part of an integer t is often abbreviated ${\displaystyle \operatorname {gcd} \left(10^{\infty },t\right).}$

   If the numbers are not to have leading zeros, then 10^k − 1 ≤ D.

A long division of 1 by 7 gives:

        0.142857...
    7 ) 1.000000
         .7
          3
          28
           2
           14
            6
            56
             4
             35
              5
              49
               1

At the last step, 1 reappears as the remainder. The cyclic remainders are {1, 3, 2, 6, 4, 5}. We rewrite the quotients with the corresponding dividend/remainders above them at all the steps:

    Dividend/Remainders    1 3 2 6 4 5
    Quotients              1 4 2 8 5 7

and also note that:

• ​¹⁄₇ = 0.142857...
• ​³⁄₇ = 0.428571...
• ​²⁄₇ = 0.285714...
• ​⁶⁄₇ = 0.857142...
• ​⁴⁄₇ = 0.571428...
• ​⁵⁄₇ = 0.714285...

By observing the remainders at each step, we can thus perform a desired cyclic permutation by multiplication. E.g.,

• The integer 142857, corresponding to remainder 1, permutes to 428571 when multiplied by 3, the corresponding remainder of the latter.
• The integer 142857, corresponding to remainder 1, permutes to 857142 when multiplied by 6, the corresponding remainder of the latter.
• The integer 857142, corresponding to remainder 6, permutes to 571428 when multiplied by ​⁵⁄₆; i.e. divided by 6 and multiplied by 5, the corresponding remainder of the latter.

In this manner, cyclical left or right shift of any number of positions can be performed.

Less importantly, this technique can be applied to any integer to shift cyclically right or left by any given number of places for the following reason:

• Every repeating decimal can be expressed as a rational number (fraction).
• Every integer, when added with a decimal point in front and concatenated with itself infinite times, can be converted to a fraction, e.g. we can transform 123456 in this manner to 0.123456123456..., which can thus be converted to fraction ​¹²³⁴⁵⁶⁄₉₉₉₉₉₉. This fraction can be further simplified but it will not be done here.
• To permute the integer 123456 to 234561, all one needs to do is to multiply 123456 by ​²³⁴⁵⁶¹⁄₁₂₃₄₅₆. This looks like cheating but if ​²³⁴⁵⁶¹⁄₁₂₃₄₅₆ is a whole number (in this case it is not), the mission is completed.

An integer X shift cyclically right by k positions when it is multiplied by an integer n. Prove its formula.

Proof

First recognize that X is the repeating digits of a repeating decimal, which always possesses cyclic behavior in multiplication. The integer X and its multiple n X then will have the following relationship:

1. The integer X is the repeating digits of the fraction ​¹⁄_F, say d_pd_p-1...d₃d₂d₁, where d_p, d_p-1, ..., d₃, d₂ and d₁ each represents a digit and p is the number of digits.
2. The multiple n X is thus the repeating digits of the fraction ​ⁿ⁄_F, say d_kd_k-1...d₃d₂d₁d_pd_p-1...d_k+2d_k+1, representing the results after right cyclical shift of k positions.
3. F must be coprime to 10 so that when ​¹⁄_F is expressed in decimal there is no preceding non-repeating digits otherwise the repeating decimal does not possess cyclic behavior in multiplication.
4. If the first remainder is taken to be n then 1 shall be the (k + 1)st remainder in the long division for ​ⁿ⁄_F in order for this cyclic permutation to take place.
5. In order that n × 10^k = 1 (mod F) then F shall be either F₀ = (n × 10^k - 1), or a factor of F₀; but excluding any values not more than n and any value having a nontrivial common factor with 10, as deduced above.

This completes the proof.

An integer X shift cyclically left by k positions when it is multiplied by an integer n. Prove its formula.

Proof

First recognize that X is the repeating digits of a repeating decimal, which always possesses a cyclic behavior in multiplication. The integer X and its multiple n X then will have the following relationship:

1. The integer X is the repeating digits of the fraction ​¹⁄_F, say d_pd_p-1...d₃d₂d₁ .
2. The multiple n X is thus the repeating digits of the fraction ​ⁿ⁄_F, say d_p-kd_p-k-1...d₃d₂d₁d_pd_p-1...d_p-k+1,

which represents the results after left cyclical shift of k positions.

1. F must be coprime to 10 so that ​¹⁄_F has no preceding non-repeating digits otherwise the repeating decimal does not possesses cyclic behavior in multiplication.
2. If the first remainder is taken to be 1 then n shall be the (k + 1)st remainder in the long division for ​¹⁄_F in order for this cyclic permutation to take place.
3. In order that 1 × 10^k = n (mode F) then F shall be either F₀ = (10^k -n), or a factor of F₀; but excluding any value not more than n, and any value having a nontrivial common factor with 10, as deduced above.

This completes the proof. The proof for non-integral multiplier such as ​ⁿ⁄_s can be derived in a similar way and is not documented here.

The permutations can be:

• Shifting right cyclically by single position (parasitic numbers);
• Shifting right cyclically by double positions;
• Shifting right cyclically by any number of positions;
• Shifting left cyclically by single position;
• Shifting left cyclically by double positions; and
• Shifting left cyclically by any number of positions

When a parasitic number is multiplied by n, not only it exhibits the cyclic behavior but the permutation is such that the last digit of the parasitic number now becomes the first digit of the multiple. For example, 102564 x 4 = 410256. Note that 102564 is the repeating digits of ​⁴⁄₃₉ and 410256 the repeating digits of ​¹⁶⁄₃₉.

An integer X shift right cyclically by double positions when it is multiplied by an integer n. X is then the repeating digits of ​¹⁄_F, whereby F = n × 10² - 1; or a factor of it; but excluding values for which ​¹⁄_F has a period length dividing 2 (or, equivalently, less than 3); and F must be coprime to 10.

Most often it is convenient to choose the smallest F that fits the above.

Problem: An integer X shift left cyclically by single position when it is multiplied by 3. Find X.

Solution: First recognize that X is the repeating digits of a repeating decimal, which always possesses some interesting cyclic behavior in multiplications. The integer X and its multiple then will have the following relationship:

• The integer X is the repeating digits of the fraction ​¹⁄_F, say ab***.
• The multiple is thus the repeating digits of the fraction ​³⁄_F, say b***a.
• In order for this cyclic permutation to take place, then 3 shall be the next remainder in the long division for ​¹⁄_F. Thus F shall be 7 as 1 × 10 ÷ 7 gives remainder 3.

This yields the results that:

X = the repeating digits of ​¹⁄₇

=142857, and

the multiple = 142857 × 3 = 428571, the repeating digits of ​³⁄₇

The other solution is represented by ​²⁄₇ x 3 = ​⁶⁄₇:

• 285714 x 3 = 857142

There are no other solutions [1] because:

• Integer n must be the subsequent remainder in a long division of a fraction ​¹⁄_F. Given that n = 10 - F, and F is coprime to 10 in order for ​¹⁄_F to be a repeating decimal, then n shall be less than 10.
• For n = 2, F must be 10 - 2 = 8. However ​¹⁄₈ does not generate a repeating decimal, similarly for n = 5.
• For n = 7, F must be 10 - 7 = 3. However 7 > 3 and ​⁷⁄₃ = 2.333 > 1 and does not fit the purpose.
• Similarly there is no solution for any other integer of n less than 10 except n = 3.

However, if the multiplier is not restricted to be an integer (though ugly), there are many other solutions from this method. E.g., if an integer X shift right cyclically by single position when it is multiplied by ​³⁄₂, then 3 shall be the next remainder after 2 in a long division of a fraction ​²⁄_F. This deduces that F = 2 x 10 - 3 = 17, giving X as the repeating digits of ​²⁄₁₇, i.e. 1176470588235294, and its multiple is 1764705882352941.

The following summarizes some of the results found in this manner:

An integer X shift left cyclically by double positions when it is multiplied by an integer n. X is then the repeating digits of ​¹⁄_F, whereby F is R = 10² - n, or a factor of R; excluding values of F for which ​¹⁄_F has a period length dividing 2 (or, equivalently, less than 3); and F must be coprime to 10.

Most often it is convenient to choose the smallest F that fits the above.

In duodecimal system, the transposable integers are: (using inverted two and three for ten and eleven, respectively)

Note that the “Shifting left cyclically by single position” problem has no solution for the multiplier less than 12 except 2 and 5, the same problem in decimal system has no solution for the multiplier less than 10 except 3.

• P. Yiu, k-right-transposable integers, k-left-transposable integers Chap.18.1, 18.2 pp. 168/360 in 'Recreational Mathematics', https://web.archive.org/web/20090901180500/http://math.fau.edu/Yiu/RecreationalMathematics2003.pdf
• C. A. Pickover, Wonders of Numbers, Chapter 28, Oxford University Press UK, 2000.
• Sloane, N. J. A. (ed.). "Sequence A092697 (For 1 <= n <= 9, a(n) = least number m such that the product n*m is obtained merely by shifting the rightmost digit of m to the left end)". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation.
• Gardner, Martin. Mathematical Circus: More Puzzles, Games, Paradoxes and Other Mathematical Entertainments From Scientific American. New York: The Mathematical Association of America, 1979. pp. 111–122.
• Kalman, Dan; 'Fractions with Cycling Digit Patterns' The College Mathematics Journal, Vol. 27, No. 2. (Mar., 1996), pp. 109–115.
• Leslie, John. "The Philosophy of Arithmetic: Exhibiting a Progressive View of the Theory and Practice of ....", Longman, Hurst, Rees, Orme, and Brown, 1820, ISBN 1-4020-1546-1
• Wells, David; "The Penguin Dictionary of Curious and Interesting Numbers", Penguin Press. ISBN 0-14-008029-5