Kaprekar's routine

Determination of Kaprekar numbers

In the following, "Kaprekar's constant k" refers to a number that becomes a positive fixed point k as a result of Kaprekar's routine.

In 1981, G. D. Prichett, et al. showed that the Kaprekar's constants in base 10 are limited to two numbers, 495 (3 digits) and 6174 (4 digits). They also classified the Kaprekar numbers into four types, but there was some overlap in the classification.

In 2005, Y. Hirata calculated all fixed points up to 31 decimal digits and examined their distribution.

In 2024, Haruo Iwasaki of the Ranzan Mathematics Study Group (headed by Kenichi Iyanaga) showed that in order for a natural number to be a Kaprekar number, it must belong to one of five mutually disjoint sets composed of combinations of the seven numbers 495, 6174, 36, 123456789, 27, 875421 and 09. Iwasaki also showed that this new classification using the five sets includes a corrected classification by Prichett, et al.

As a result, if n is considered as a constant, then the number of decimal n-digit Kaprekar numbers is determined by two types of equations:

(1) ${\displaystyle n=3x(x\ge 1),}$ ......... For the sequence of x 3-digit constants 495

(2) ${\displaystyle n=4+2x(x\ge 0),}$ ...... Sequence of 4-digit constant 6174 followed by x 2-digit constants 36

or by three types of Diophantine equations:

(3) ${\displaystyle n=9x+2y(x\ge 1,y\ge 0),}$ ......... Sequence of x 123456789's and y 36's

(4) ${\displaystyle n=9x+14y(x\ge 1,y\ge 1),}$ ...... Sequence of x 123456789's and y (36 495495 272727)s

(5) ${\displaystyle n=6x+2y+9z+2u(x\ge 1,y\ge 1,z\ge 0,u\ge 0).}$ ... Sequence of 124578's, 09's, 123456789's and 36's

It was found that the number of integer solutions (sets of x~u) of the equations that can be established is the same as the number of solutions that express all of the n-digit Kaprekar numbers.

The above equations confirm that there are no other Kaprekar's constants than 495 and 6174. There are no Kaprekar numbers for 1, 2, 5, or 7 digits, since they do not satisfy any of equations (1) through (5). For six-digit numbers, there are two solutions that satisfy equations (1) and (2). Furthermore, it is clear that even-digits with greater than or equal to 8, and with 9 digits, or odd-digits with greater than or equal to 15 digits have multiple solutions. Although 11-digit and 13-digit numbers have only one solution, it forms a loop of five numbers and a loop of two numbers, respectively. Hence, Prichett's result that the Kaprekar's constants are limited to 495 (3 digits) and 6174 (4 digits) is verified.

Therefore, the problem of determining all of the Kaprekar's constants and the number of these was solved. An example below will explain the Iwasaki's result.

Example: In the case where decimal digits n = 23, since n is an odd number and is not a multiple of 3, the equations (1) and (2) do not hold, and the only equations that can hold are (3), (4) and (5). If the operation (denoted by K₁₀ ) defined above is applied once to the numbers corresponding to the solutions of these equations, seven Kaprekar numbers can be obtained.

(3) The solution to 23 = 9x + 2y is

(x, y) = (1, 7) : ...... Sequence of a 123456789 followed by seven 36's

K₁₀ (123456789 36363636363636) = 86433333331976666666532.

(4) The solution to 23 = 9x + 14y is

(x, y) = (1, 1) : ...... Sequence of 123456789 followed by 36 495495 272727

K₁₀ (123456789 36 495495 272727) = 87765443219997765543222.

(5) The solutions to 23 = 6x + 2y + 9z + 2u are

(x, y, z, u) = (1, 4, 1, 0) : ...... Sequence of 124578, four 09's and 123456789

K₁₀ (124578 09090909 123456789) = 99998765420987543210001,

(x, y, z, u) = (1, 3, 1, 1) : ...... Sequence of 124578, three 09's, 123456789 and 36

K₁₀ (124578 090909 123456789 36) = 99987654320987654321001,

(x, y, z, u) = (1, 2, 1, 2) : ...... Sequence of 124578, two 09's, 123456789 and two 36's

K₁₀ (124578 0909 123456789 3636) = 99876543320987665432101,

(x, y, z, u) = (1, 1, 1, 3) : ...... Sequence of 124578, 09, 123456789 and three 36's

K₁₀ (124578 09 123456789 363636) = 98765433320987666543211, and

(x, y, z, u) = (2, 1, 1, 0) : ...... Sequence of two 124578's, 09 and 123456789

K₁₀ (124578124578 09 123456789) = 98776554210988754432211.

Families of Kaprekar's constants

In case where even base (b = 2k)

It can be shown that all natural numbers

${\displaystyle m=(k)b^{2n+3}\left(\sum _{i=0}^{n-1}{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b\left(\sum _{i=0}^{n-1}{b}^i\right)+(k)}$

are fixed points of the Kaprekar mapping in even base b = 2k for all natural numbers n.

Proof

${\displaystyle \alpha =(2k-1)b^{2n+2}\left(\sum _{i=0}^n{b}^i\right)+(k)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+(k-1)\left(\sum _{i=0}^n{b}^i\right)}$

${\displaystyle \beta =(k-1)b^{2n+2}\left(\sum _{i=0}^n{b}^i\right)+(k)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+(2k-1)\left(\sum _{i=0}^n{b}^i\right)}$

${\displaystyle \begin{array}{cc}{K}_b(m)& =\alpha -\beta \\ & =((2k-1)-(k-1))b^{2n+2}\left(\sum _{i=0}^n{b}^i\right)+(k-k)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+((k-1)-(2k-1))\left(\sum _{i=0}^n{b}^i\right)\\ & =k{b}^{2n+2}\left(\sum _{i=0}^n{b}^i\right)-k\left(\sum _{i=0}^n{b}^i\right)\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+b^{2n+2}-k\left(\sum _{i=0}^n{b}^i\right)\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k)b^{2n+1}-k\left(\sum _{i=0}^n{b}^i\right)\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{2n+1}+b^{2n+1}-k\left(\sum _{i=0}^n{b}^i\right)\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{2n+1-1}\left(\sum _{i=0}^1{b}^i\right)+b^{2n+1-1}-k\left(\sum _{i=0}^n{b}^i\right)\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{2n+1-n}\left(\sum _{i=0}^n{b}^i\right)+b^{2n+1-n}-k\left(\sum _{i=0}^n{b}^i\right)\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+b^{n+1}-k\left(\sum _{i=0}^n{b}^i\right)\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+(2k)b^n-k\left(\sum _{i=0}^n{b}^i\right)\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+k{b}^n-k\left(\sum _{i=0}^{n-1}{b}^i\right)\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{n+1-1}+b^{n+1-1}-k\left(\sum _{i=0}^{n-n}{b}^i\right)\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{n+1-n}\left(\sum _{i=0}^n{b}^i\right)+b^{n+1-n}-k\left(\sum _{i=0}^{n-n}{b}^i\right)\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b\left(\sum _{i=0}^n{b}^i\right)+b-k\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b\left(\sum _{i=0}^n{b}^i\right)+2k-k\\ & =k{b}^{2n+3}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b^{2n+2}+(2k-1)b^{n+1}\left(\sum _{i=0}^n{b}^i\right)+(k-1)b\left(\sum _{i=0}^n{b}^i\right)+k\\ & =m\end{array}}$

Notes and citations