Transposable integer explained

The digits of some specific integers permute or shift cyclically when they are multiplied by a number n. Examples are:

These specific integers, known as 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. = 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,

m-1\right),
\gcd\left(N,10
c,10

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:

Nc=10N-d\left(10m-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:

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

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:

  1. An integer X shift right cyclically by k positions when it is multiplied by an integer n. X is then the repeating digits of, 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.
  2. An integer X shift left cyclically by k positions when it is multiplied by an integer n. X is then the repeating digits of, 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 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 < 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

10p-1
F

where p is a period length of ; 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 >, i.e. j > .

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 . X is then the repeating digits of, 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 < 1. Again it is convenient to choose the smallest F that fits the above.

The solutions can be expressed by the formula:

X=js

10p-1
F

where p is defined likewise; and F is made coprime to 10 by the same process as before.

To exclude integers that begin with zeros from the solutions, select an integer j such that >, i.e. j > .

Again if > 1, there is no solution.

Direct representation

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

X=D

10m-1
n10k-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 < 10k.

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

X=D

10m-1
10k-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:

D<10k
n

-1,

  1. and the 10-part (the product of the terms corresponding to the primes 2 and 5 of the factorization) of 10k - n divides D.

The 10-part of an integer t is often abbreviated

\operatorname{gcd}\left(10infty,t\right).

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

Cyclic permutation by multiplication

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 . 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:

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

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:

Proof of formula for cyclical right shift operation

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, say dpdp-1...d3d2d1, where dp, dp-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, say dkdk-1...d3d2d1dpdp-1...dk+2dk+1, representing the results after right cyclical shift of k positions.
  3. F must be coprime to 10 so that when 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 in order for this cyclic permutation to take place.
  5. In order that n × 10k = 1 (mod F) then F shall be either F0 = (n × 10k - 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.

Proof of formula for cyclical left shift operation

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, say dpdp-1...d3d2d1 .
  2. The multiple n X is thus the repeating digits of the fraction, say dp-kdp-k-1...d3d2d1dpdp-1...dp-k+1,

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

  1. F must be coprime to 10 so that 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 in order for this cyclic permutation to take place.
  3. In order that 1 × 10k = n (mode F) then F shall be either F0 = (10k -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 can be derived in a similar way and is not documented here.

Shifting an integer cyclically

The permutations can be:

Parasitic numbers

See main article: parasitic number.

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 and 410256 the repeating digits of .

Shifting right cyclically by double positions

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

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

Summary of results

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:

Multiplier nSolutionRepresented byOther Solutions
20050251256 2814070351 7587939698 4924623115 5778894472 3618090452 2613065326 6331658291 4572864321 608040201 x 2 = period = 99i.e. 99 repeating digits.,, ...,
30033444816 0535117056 8561872909 6989966555 1839464882 9431438127 090301 x 3 = period = 66

299 = 13×23

,, ..., some special cases are illustrated below
3076923 x 3 = period = 6,,
30434782608 6956521739 13 x 3 = period = 22,, ...,
40025062656 64160401 x 4 = period = 18

399 = 3×7×19

,, ..., some special cases are illustrated below
4142857 x 4 = period = 6-
40526315789 47368421 x 4 = period = 18,,
5(a cyclic number with a period of 498) x 5 = 499 is a full reptend prime,, ...,

Note that:

There are many other possibilities.

Shifting left cyclically by single position

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:

This yields the results that:

X = the repeating digits of

=142857, and

the multiple = 142857 × 3 = 428571, the repeating digits of

The other solution is represented by x 3 = :

There are no other solutions [1] because:

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, then 3 shall be the next remainder after 2 in a long division of a fraction . This deduces that F = 2 x 10 - 3 = 17, giving X as the repeating digits of, i.e. 1176470588235294, and its multiple is 1764705882352941.

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

Multiplier SolutionRepresented byOther Solutions
105263157894736842 × = A 2-parasitic numberOther 2-parasitic numbers:,,,,,,,
1176470588235294 × = ,,,
153846 × = -
18 × = -
1304347826086956521739 × = ,,,,,
190476 × = -

Shifting left cyclically by double positions

An integer X shift left cyclically by double positions when it is multiplied by an integer n. X is then the repeating digits of, whereby is = 102 - n, or a factor of ; excluding values of for which 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 that fits the above.

Summary of results

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:

Multiplier nSolutionRepresented byOther Solutions
2142857 × 2 = ,
30103092783 5051546391 7525773195 8762886597 9381443298 9690721649 4845360824 7422680412 3711340206 185567 x 3 = ,,,, ....,,
4No solution--
50526315789 47368421 x 5 = ,
60212765957 4468085106 3829787234 0425531914 893617 x 6 = ,,,,,
70322580645 16129 x 7 = ,,,,,,,,,,
80434782608 6956521739 13 x 8 =
9076923 x 9 = ,,,,,,,,
10No solution--
110112359550 5617977528 0898876404 4943820224 7191 x 11 = ,,,,,,
12No solution--
130344827586 2068965517 24137931 x 13 = ,,,,
140232558139 5348837209 3 x 14 = ,
150588235294 117647 x 15 = -

Other bases

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

Multiplier nSmallest solution such that the multiplication moves the last digit to leftDigitsRepresented bySmallest solution such that the multiplication moves the first digit to rightDigitsRepresented by
206316948421Ɛ x 2 = 24974 x 2 =
324974 x 3 = no solution
40309236ᘔ8820 61647195441 x 4 = no solution
5025355ᘔ94330 73ᘔ458409919 Ɛ715125 x 5 = 186ᘔ356 x 5 =
6020408142854 ᘔ997732650ᘔ1 83469163061 x 6 = no solution
701899Ɛ864406 Ɛ33ᘔᘔ1542391 374594930525 5Ɛ17135 x 7 = no solution
8076Ɛ456 x 8 = no solution
9014196486344 59Ɛ9384Ɛ26Ɛ5 33040547216ᘔ 1155Ɛ3Ɛ12978 ᘔ399145 x 9 = no solution
08579214Ɛ364 29ᘔ714 x ᘔ = no solution
Ɛ011235930336 ᘔ53909ᘔ873Ɛ3 25819Ɛ997505 5Ɛ54ᘔ3145ᘔ42 694157078404 491Ɛ155 x Ɛ = no solution

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.

Notes

  1. P. Yiu, k-right-transposable integers, Chap.18.1 'Recreational Mathematics'

References