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Inmathematics, thetransposable integers are integers thatpermute orshift cyclically when they are multiplied by another integer${\displaystyle n}$. 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 alwayscyclic numbers. The characterization of such numbers can be done usingrepeating decimals (and thus the related fractions), or directly.

For anyinteger 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 withvinculum on top is adequate, the intention of the above expression is to show that the sixcyclic 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.

Thegreatest common divisor (gcd) between any cyclic permutation of anm-digit integer and 10^m − 1 is constant. Expressed as a formula,

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

whereN is anm-digit integer; andN_c is any cyclic permutation ofN.

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)

IfN is anm-digit integer, the numberN_c, obtained by shiftingN to the left cyclically, can be obtained from:

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

whered is the first digit ofN andm is the number of digits.

This explains the above common gcd and the phenomenon is true in anybase if 10 is replaced byb, 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, and509157⁄999999.

Reduced to their lowest terms using the common gcd, they are:

• 25⁄273,250⁄273,43⁄273,157⁄273,205⁄273, and139⁄273.

That is, these fractions when expressedin lowest terms, have the same denominator. This is true for cyclic permutations of any integer.

An integral multiplier refers to the multipliern being an integer:

1. An integerX shiftright cyclically byk positions when it is multiplied byan integern.X is then the repeating digits of1⁄F, wherebyF isF₀ =n 10^k − 1 (F₀ iscoprime to 10), or a factor ofF₀; excluding any values ofF which are not more thann.
2. An integerX shiftleft cyclically byk positions when it is multiplied byan integern.X is then the repeating digits of1⁄F, wherebyF isF₀ = 10^k -n, or a factor ofF₀; excluding any values ofF which are not more thann and which are notcoprime to 10.

It is necessary for F to be coprime to 10 in order that1⁄F is a repeating decimal without any preceding non-repeating digits (see multiple sections ofRepeating decimal). If there are digits not in a period, then there is no corresponding solution.

For these two cases, multiples ofX, i.e. (j X) are also solutions provided that the integeri satisfies the conditionn j⁄F < 1. Most often it is convenient to choose the smallestF that fits the above. The solutions can be expressed by the formula:

${\displaystyle X=j\frac{{10}^p-1}{F}}$

wherep is a period length of1⁄F; andF is a factor ofF₀ coprime to 10.

E.g,F₀ = 1260 = 2² × 3² × 5 × 7. The factors excluding 2 and 5 recompose toF = 3² × 7 = 63. Alternatively, strike off all theending 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 alsoF = 63.

To exclude integers that begin with zeros from the solutions, select an integerj such thatj⁄F >1⁄10, i.e.j >F⁄10.

There is no solution whenn >F.

An integerX shiftleft cyclically byk positions when it is multiplied bya fractionn⁄s.X is then the repeating digits ofs⁄F, wherebyF isF₀ = s 10^k -n, or a factor ofF₀; andF must be coprime to 10.

For this third case, multiples ofX, i.e. (j X) are again solutions but the condition to be satisfied for integerj is thatn j⁄F < 1. Again it is convenient to choose the smallestF that fits the above.

The solutions can be expressed by the formula:

${\displaystyle X=js\frac{{10}^p-1}{F}}$

wherep is defined likewise; andF is made coprime to 10 by the same process as before.

To exclude integers that begin with zeros from the solutions, select an integerj such thatj s⁄F >1⁄10, i.e.j >F⁄10s.

Again ifj s⁄F > 1, there is no solution.

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

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

   wherem is the number of digits ofX, andD, thek-digit number shifted from the low end ofX to the high end ofnX, satisfiesD < 10^k.

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

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

   wherem is the number of digits ofX, andD, thek-digit number shifted from the high end ofX to the low end ofnX, satisfies:

   1. 
   2. and the 10-part (the product of the terms corresponding to the primes 2 and 5 of thefactorization) of 10^k − n dividesD.

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

   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          328           214            656             435              549               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 desiredcyclic 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 by5⁄6; 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 toshift 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 fraction123456⁄999999. 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 by234561⁄123456. This looks like cheating but if234561⁄123456 is a whole number (in this case it is not), the mission is completed.

An integerX shift cyclically right byk positions when it is multiplied by an integern. Prove its formula.

Proof

First recognize thatX is the repeating digits of arepeating decimal, which always possesses cyclic behavior in multiplication. The integerX and its multiplen X then will have the following relationship:

1. The integerX is the repeating digits of the fraction1⁄F, sayd_pd_p-1...d₃d₂d₁, whered_p,d_p-1, ...,d₃,d₂ andd₁ each represents a digit andp is the number of digits.
2. The multiplen X is thus the repeating digits of the fractionn⁄F, sayd_kd_k-1...d₃d₂d₁d_pd_p-1...d_k+2d_k+1, representing the results after right cyclical shift ofk positions.
3. F must be coprime to 10 so that when1⁄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 ben then1 shall be the (k + 1)st remainder in the long division forn⁄F in order for this cyclic permutation to take place.
5. In order thatn × 10^k = 1 (modF) thenF shall be eitherF₀ = (n × 10^k - 1), or a factor ofF₀; but excluding any values not more thann and any value having a nontrivial common factor with 10, as deduced above.

This completes the proof.

An integerX shift cyclically left byk positions when it is multiplied byan integern. Prove its formula.

Proof

First recognize thatX is the repeating digits of arepeating decimal, which always possesses a cyclic behavior in multiplication. The integerX and its multiplen X then will have the following relationship:

1. The integerX is the repeating digits of the fraction1⁄F, sayd_pd_p-1...d₃d₂d₁ .
2. The multiplen X is thus the repeating digits of the fractionn⁄F, sayd_p-kd_p-k-1...d₃d₂d₁d_pd_p-1...d_p-k+1,

which represents the results after left cyclical shift ofk positions.

1. F must be coprime to 10 so that1⁄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 thenn shall be the (k + 1)st remainder in the long division for1⁄F in order for this cyclic permutation to take place.
3. In order that 1 × 10^k =n (modeF) thenF shall be eitherF₀ = (10^k -n), or a factor ofF₀; but excluding any value not more thann, and any value having a nontrivial common factor with 10, as deduced above.

This completes the proof. The proof for non-integral multiplier such asn⁄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 of4⁄39 and 410256 the repeating digits of16⁄39.

An integerX shift right cyclically by double positions when it is multiplied by an integern.X is then the repeating digits of1⁄F, wherebyF =n × 10² - 1; or a factor of it; but excluding values for which1⁄F has a period length dividing 2 (or, equivalently, less than 3); andF must be coprime to 10.

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

The following multiplication moves the last two digits of each original integer to the first two digits and shift every other digits to the right:

Note that:

• 299 = 13 x 23, and the period of1⁄299 is accurately determined by the formula, LCM(6, 22) = 66, according toRepeating decimal#Generalization.
• 399 = 3 x 7 x 19, and the period of1⁄399 is accurately determined by the formula, LCM(1, 6, 18) = 18.

There are many other possibilities.

Problem: An integerX shift left cyclically by single position when it is multiplied by 3. FindX.

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

• The integerX is the repeating digits of the fraction1⁄F, sayab***.
• The multiple is thus the repeating digits of the fraction3⁄F, sayb***a.
• In order for this cyclic permutation to take place, then 3 shall be the next remainder in the long division for1⁄F. ThusF shall be 7 as 1 × 10 ÷ 7 gives remainder 3.

This yields the results that:

X = the repeating digits of1⁄7

=142857, and

the multiple = 142857 × 3 = 428571, the repeating digits of3⁄7

The other solution is represented by2⁄7 x 3 =6⁄7:

• 285714 x 3 = 857142

There are no other solutions^[1] because:

• Integern must be the subsequent remainder in a long division of a fraction1⁄F. Given that n = 10 - F, and F is coprime to 10 in order for1⁄F to be a repeating decimal, thenn shall be less than 10.
• Forn = 2, F must be 10 - 2 = 8. However1⁄8 does not generate a repeating decimal, similarly forn = 5.
• Forn = 7, F must be 10 - 7 = 3. However 7 > 3 and7⁄3 = 2.333 > 1 and does not fit the purpose.
• Similarly there is no solution for any other integer ofn less than 10 exceptn = 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 integerX shift right cyclically by single position when it is multiplied by3⁄2, then 3 shall be the next remainder after 2 in a long division of a fraction2⁄F. This deduces that F = 2 x 10 - 3 = 17, givingX as the repeating digits of2⁄17, i.e. 1176470588235294, and its multiple is 1764705882352941.

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

An integerX shift left cyclically by double positions when it is multiplied by an integern.X is then the repeating digits of1⁄F, wherebyF isR = 10² - n, or a factor ofR; excluding values ofF for which1⁄F has a period length dividing 2 (or, equivalently, less than 3); andF must be coprime to 10.

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

The following summarizes some of the results obtained in this manner, where the white spaces between the digits divide the digits into 10-digit groups:

Induodecimal 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)".TheOn-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

• Base-dependent integer sequences

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