Kaprekar number

Short description: Base-dependent property of integers

In mathematics, a natural number in a given number base is a [math]\displaystyle{ p }[/math]-Kaprekar number if the representation of its square in that base can be split into two parts, where the second part has [math]\displaystyle{ p }[/math] digits, that add up to the original number. For example, in base 10, 45 is a 2-Kaprekar number, because 45² = 2025, and 20 + 25 = 45. The numbers are named after D. R. Kaprekar.

Definition and properties

Let [math]\displaystyle{ n }[/math] be a natural number. We define the Kaprekar function for base [math]\displaystyle{ b \gt 1 }[/math] and power [math]\displaystyle{ p \gt 0 }[/math] [math]\displaystyle{ F_{p, b} : \mathbb{N} \rightarrow \mathbb{N} }[/math] to be the following:

[math]\displaystyle{ F_{p, b}(n) = \alpha + \beta }[/math],

where [math]\displaystyle{ \beta = n^2 \bmod b^p }[/math] and

[math]\displaystyle{ \alpha = \frac{n^2 - \beta}{b^p} }[/math]

A natural number [math]\displaystyle{ n }[/math] is a [math]\displaystyle{ p }[/math]-Kaprekar number if it is a fixed point for [math]\displaystyle{ F_{p, b} }[/math], which occurs if [math]\displaystyle{ F_{p, b}(n) = n }[/math]. [math]\displaystyle{ 0 }[/math] and [math]\displaystyle{ 1 }[/math] are trivial Kaprekar numbers for all [math]\displaystyle{ b }[/math] and [math]\displaystyle{ p }[/math], all other Kaprekar numbers are nontrivial Kaprekar numbers.

The earlier example of 45 satisfies this definition with [math]\displaystyle{ b = 10 }[/math] and [math]\displaystyle{ p = 2 }[/math], because

[math]\displaystyle{ \beta = n^2 \bmod b^p = 45^2 \bmod 10^2 = 25 }[/math]

[math]\displaystyle{ \alpha = \frac{n^2 - \beta}{b^p} = \frac{45^2 - 25}{10^2} = 20 }[/math]

[math]\displaystyle{ F_{2, 10}(45) = \alpha + \beta = 20 + 25 = 45 }[/math]

A natural number [math]\displaystyle{ n }[/math] is a sociable Kaprekar number if it is a periodic point for [math]\displaystyle{ F_{p, b} }[/math], where [math]\displaystyle{ F_{p, b}^k(n) = n }[/math] for a positive integer [math]\displaystyle{ k }[/math] (where [math]\displaystyle{ F_{p, b}^k }[/math] is the [math]\displaystyle{ k }[/math]th iterate of [math]\displaystyle{ F_{p, b} }[/math]), and forms a cycle of period [math]\displaystyle{ k }[/math]. A Kaprekar number is a sociable Kaprekar number with [math]\displaystyle{ k = 1 }[/math], and a amicable Kaprekar number is a sociable Kaprekar number with [math]\displaystyle{ k = 2 }[/math].

The number of iterations [math]\displaystyle{ i }[/math] needed for [math]\displaystyle{ F_{p, b}^{i}(n) }[/math] to reach a fixed point is the Kaprekar function's persistence of [math]\displaystyle{ n }[/math], and undefined if it never reaches a fixed point.

There are only a finite number of [math]\displaystyle{ p }[/math]-Kaprekar numbers and cycles for a given base [math]\displaystyle{ b }[/math], because if [math]\displaystyle{ n = b^p + m }[/math], where [math]\displaystyle{ m \gt 0 }[/math] then

[math]\displaystyle{ \begin{align} n^2 & = (b^p + m)^2 \\ & = b^{2p} + 2mb^p + m^2 \\ & = (b^p + 2m)b^p + m^2 \\ \end{align} }[/math]

and [math]\displaystyle{ \beta = m^2 }[/math], [math]\displaystyle{ \alpha = b^p + 2m }[/math], and [math]\displaystyle{ F_{p, b}(n) = b^p + 2m + m^2 = n + (m^2 + m) \gt n }[/math]. Only when [math]\displaystyle{ n \leq b^p }[/math] do Kaprekar numbers and cycles exist.

If [math]\displaystyle{ d }[/math] is any divisor of [math]\displaystyle{ p }[/math], then [math]\displaystyle{ n }[/math] is also a [math]\displaystyle{ p }[/math]-Kaprekar number for base [math]\displaystyle{ b^p }[/math].

In base [math]\displaystyle{ b = 2 }[/math], all even perfect numbers are Kaprekar numbers. More generally, any numbers of the form [math]\displaystyle{ 2^n (2^{n + 1} - 1) }[/math] or [math]\displaystyle{ 2^n (2^{n + 1} + 1) }[/math] for natural number [math]\displaystyle{ n }[/math] are Kaprekar numbers in base 2.

Set-theoretic definition and unitary divisors

We can define the set [math]\displaystyle{ K(N) }[/math] for a given integer [math]\displaystyle{ N }[/math] as the set of integers [math]\displaystyle{ X }[/math] for which there exist natural numbers [math]\displaystyle{ A }[/math] and [math]\displaystyle{ B }[/math] satisfying the Diophantine equation

[math]\displaystyle{ X^2 = AN + B }[/math], where [math]\displaystyle{ 0 \leq B \lt N }[/math]

[math]\displaystyle{ X = A + B }[/math]

An [math]\displaystyle{ n }[/math]-Kaprekar number for base [math]\displaystyle{ b }[/math] is then one which lies in the set [math]\displaystyle{ K(b^n) }[/math].

It was shown in 2000 that there is a bijection between the unitary divisors of [math]\displaystyle{ N - 1 }[/math] and the set [math]\displaystyle{ K(N) }[/math] defined above. Let [math]\displaystyle{ \operatorname{Inv}(a, c) }[/math] denote the multiplicative inverse of [math]\displaystyle{ a }[/math] modulo [math]\displaystyle{ c }[/math], namely the least positive integer [math]\displaystyle{ m }[/math] such that [math]\displaystyle{ am = 1 \bmod c }[/math], and for each unitary divisor [math]\displaystyle{ d }[/math] of [math]\displaystyle{ N - 1 }[/math] let [math]\displaystyle{ e = \frac{N - 1}{d} }[/math] and [math]\displaystyle{ \zeta(d) = d\ \text{Inv}(d, e) }[/math]. Then the function [math]\displaystyle{ \zeta }[/math] is a bijection from the set of unitary divisors of [math]\displaystyle{ N - 1 }[/math] onto the set [math]\displaystyle{ K(N) }[/math]. In particular, a number [math]\displaystyle{ X }[/math] is in the set [math]\displaystyle{ K(N) }[/math] if and only if [math]\displaystyle{ X = d\ \text{Inv}(d, e) }[/math] for some unitary divisor [math]\displaystyle{ d }[/math] of [math]\displaystyle{ N - 1 }[/math].

The numbers in [math]\displaystyle{ K(N) }[/math] occur in complementary pairs, [math]\displaystyle{ X }[/math] and [math]\displaystyle{ N - X }[/math]. If [math]\displaystyle{ d }[/math] is a unitary divisor of [math]\displaystyle{ N - 1 }[/math] then so is [math]\displaystyle{ e = \frac{N - 1}{d} }[/math], and if [math]\displaystyle{ X = d\operatorname{Inv}(d, e) }[/math] then [math]\displaystyle{ N - X = e\operatorname{Inv}(e, d) }[/math].

Kaprekar numbers for [math]\displaystyle{ F_{p, b} }[/math]

b = 4k + 3 and p = 2n + 1

Let [math]\displaystyle{ k }[/math] and [math]\displaystyle{ n }[/math] be natural numbers, the number base [math]\displaystyle{ b = 4k + 3 = 2(2k + 1) + 1 }[/math], and [math]\displaystyle{ p = 2n + 1 }[/math]. Then:

• [math]\displaystyle{ X_1 = \frac{b^p - 1}{2} = (2k + 1) \sum_{i = 0}^{p - 1} b^i }[/math] is a Kaprekar number.

Proof

Let

[math]\displaystyle{ \begin{align} X_1 & = \frac{b^p - 1}{2} \\ & = \frac{b - 1}{2} \sum_{i = 0}^{p - 1} b^i \\ & = \frac{4k + 3 - 1}{2} \sum_{i = 0}^{2n + 1 - 1} b^i \\ & = (2k + 1) \sum_{i = 0}^{2n} b^i \end{align} }[/math]

Then,

[math]\displaystyle{ \begin{align} X_1^2 & = \left(\frac{b^p - 1}{2}\right)^2 \\ & = \frac{b^{2p} - 2b^p + 1}{4} \\ & = \frac{b^p(b^p - 2) + 1}{4} \\ & = \frac{(4k + 3)^{2n + 1}(b^p - 2) + 1}{4} \\ & = \frac{(4k + 3)^{2n}(b^p - 2)(4k + 4) - (4k + 3)^{2n}(b^p - 2) + 1}{4} \\ & = \frac{-(4k + 3)^{2n}(b^p - 2) + 1}{4} + (k + 1)(4k + 3)^{2n}(b^p - 2) \\ & = \frac{-(4k + 3)^{2n - 1}(b^p - 2)(4k + 4) + (4k + 3)^{2n - 1}(b^p - 2) + 1}{4} + (k + 1)b^{2n}(b^{2n + 1} - 2) \\ & = \frac{(4k + 3)^{2n - 1}(b^p - 2) + 1}{4} + (k + 1)b^{2n}(b^p - 2) - (k + 1)b^{2n - 1}(b^{2n + 1} - 2) \\ & = \frac{(4k + 3)^{p - 2}(b^p - 2) + 1}{4} + \sum_{i = p - 2}^{p - 1} (-1)^i(k + 1)b^i(b^p - 2) \\ & = \frac{(4k + 3)^{p - 2}(b^p - 2) + 1}{4} + (b^p - 2)(k + 1)\sum_{i = p - 2}^{p - 1} (-1)^i b^i \\ & = \frac{(4k + 3)^{1}(b^p - 2) + 1}{4} + (b^p - 2)(k + 1)\sum_{i = 1}^{p - 1} (-1)^i b^i \\ & = \frac{-(b^p - 2) + 1}{4} + (b^p - 2)(k + 1)\sum_{i = 0}^{p - 1} (-1)^i b^i \\ & = (b^p - 2)(k + 1)\left(\sum_{i = 0}^{2n} (-1)^i b^i\right) + \frac{-b^{2n + 1} + 3}{4} \\ & = (b^p - 2)(k + 1)\left(\sum_{i = 0}^{2n} (-1)^i b^i\right) + \frac{-4b^{2n + 1} + 3b^{2n + 1} + 3}{4} \\ & = (b^p - 2)(k + 1)\left(\sum_{i = 0}^{2n} (-1)^i b^i\right) - b^p + \frac{3b^{2n + 1} + 3}{4} \\ & = (b^p - 2)(k + 1)\left(\sum_{i = 0}^{2n} (-1)^i b^i\right) - b^p + \frac{3(4k + 3)^{p - 2} + 3}{4} + 3(k + 1) \sum_{i = p - 2}^{p - 1} (-1)^i b^i \\ & = (b^p - 2)(k + 1)\left(\sum_{i = 0}^{2n} (-1)^i b^i\right) - b^p + \frac{3(4k + 3)^{1} + 3}{4} + 3(k + 1) \sum_{i = 1}^{p - 1} (-1)^i b^i \\ & = (b^p - 2)(k + 1)\left(\sum_{i = 0}^{2n} (-1)^i b^i\right) - b^p + \frac{-3 + 3}{4} + 3(k + 1) \sum_{i = 0}^{p - 1} (-1)^i b^i \\ & = (b^p - 2)(k + 1)\left(\sum_{i = 0}^{2n} (-1)^i b^i\right) + 3(k + 1)\left(\sum_{i = 0}^{2n} (-1)^i b^i\right) - b^p \\ & = (b^p - 2 + 3)(k + 1)\left(\sum_{i = 0}^{2n} (-1)^i b^i\right) - b^p \\ & = (b^p + 1)(k + 1)\left(\sum_{i = 0}^{2n} (-1)^i b^i\right) - b^p \\ & = (b^p + 1)\left(-1 + (k + 1)\sum_{i = 0}^{2n} (-1)^i b^i\right) + 1 \\ & = (b^p + 1)\left(k + (k + 1)\sum_{i = 1}^{2n} (-1)^i b^i\right) + 1 \\ & = (b^p + 1)\left(k + (k + 1)\sum_{i = 1}^{n} b^{2i} - b^{2i - 1}\right) + 1 \\ & = (b^p + 1)\left(k + (k + 1)\sum_{i = 1}^{n} (b - 1)b^{2i - 1}\right) + 1 \\ & = (b^p + 1)\left(k + \sum_{i = 1}^{n} ((k + 1)b - k - 1)b^{2i - 1}\right) + 1 \\ & = (b^p + 1)\left(k + \sum_{i = 1}^{n} (kb + (4k + 3) - k - 1)b^{2i - 1}\right) + 1 \\ & = (b^p + 1)\left(k + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1}\right) + 1 \\ & = b^p \left(k + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1}\right) + \left(k + 1 + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1}\right) \end{align} }[/math]

The two numbers [math]\displaystyle{ \alpha }[/math] and [math]\displaystyle{ \beta }[/math] are

[math]\displaystyle{ \beta = X_1^2 \bmod b^p = k + 1 + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1} }[/math]

[math]\displaystyle{ \alpha = \frac{X_1^2 - \beta}{b^p} = k + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1} }[/math]

and their sum is

[math]\displaystyle{ \begin{align} \alpha + \beta & = \left(k + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1}\right) + \left(k + 1 + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1}\right) \\ & = 2k + 1 + \sum_{i = 1}^{n} ((2k)b + 2(3k + 2))b^{2i - 1} \\ & = 2k + 1 + \sum_{i = 1}^{n} ((2k)b + (6k + 4))b^{2i - 1} \\ & = 2k + 1 + \sum_{i = 1}^{n} ((2k)b + (4k + 3))b^{2i - 1} + (2k + 1)b^{2i - 1} \\ & = 2k + 1 + \sum_{i = 1}^{n} ((2k + 1)b)b^{2i - 1} + (2k + 1)b^{2i - 1} \\ & = 2k + 1 + \sum_{i = 1}^{n} (2k + 1)b^{2i} + (2k + 1)b^{2i - 1} \\ & = 2k + 1 + \sum_{i = 1}^{2n} (2k + 1)b^{i} \\ & = \sum_{i = 0}^{2n} (2k + 1)b^{i} \\ & = (2k + 1) \sum_{i = 0}^{2n} b^i & = X_1 \\ \end{align} }[/math]

Thus, [math]\displaystyle{ X_1 }[/math] is a Kaprekar number.

• [math]\displaystyle{ X_2 = \frac{b^p + 1}{2} = X_1 + 1 }[/math] is a Kaprekar number for all natural numbers [math]\displaystyle{ n }[/math].

Proof

Let

[math]\displaystyle{ \begin{align} X_2 & = \frac{b^{2n + 1} + 1}{2} \\ & = \frac{b^{2n + 1} - 1}{2} + 1 \\ & = X_1 + 1 \end{align} }[/math]

Then,

[math]\displaystyle{ \begin{align} X_2^2 & = (X_1 + 1)^2 \\ & = X_1^2 + 2 X_1 + 1 \\ & = X_1^2 + 2 X_1 + 1 \\ & = b^p \left(k + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1}\right) + \left(k + 1 + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1}\right) + b^p - 1 + 1 \\ & = b^p \left(k + 1 + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1}\right) + \left(k + 1 + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1}\right) \end{align} }[/math]

The two numbers [math]\displaystyle{ \alpha }[/math] and [math]\displaystyle{ \beta }[/math] are

[math]\displaystyle{ \beta = X_2^2 \bmod b^p = k + 1 + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1} }[/math]

[math]\displaystyle{ \alpha = \frac{X_2^2 - \beta}{b^p} = k + 1 + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1} }[/math]

and their sum is

[math]\displaystyle{ \begin{align} \alpha + \beta & = \left(k + 1 + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1}\right) + \left(k + 1 + \sum_{i = 1}^{n} (kb + (3k + 2))b^{2i - 1}\right) \\ & = 2k + 2 + \sum_{i = 1}^{n} ((2k)b + 2(3k + 2))b^{2i - 1} \\ & = 2k + 2 + \sum_{i = 1}^{n} ((2k)b + (6k + 4))b^{2i - 1} \\ & = 2k + 2 + \sum_{i = 1}^{n} ((2k)b + (4k + 3))b^{2i - 1} + (2k + 1)b^{2i - 1} \\ & = 2k + 2 + \sum_{i = 1}^{n} ((2k + 1)b)b^{2i - 1} + (2k + 1)b^{2i - 1} \\ & = 2k + 2 + \sum_{i = 1}^{n} (2k + 1)b^{2i} + (2k + 1)b^{2i - 1} \\ & = 2k + 2 + \sum_{i = 1}^{2n} (2k + 1)b^{i} \\ & = 1 + \sum_{i = 0}^{2n} (2k + 1)b^{i} \\ & = 1 + (2k + 1) \sum_{i = 0}^{2n} b^{i} \\ & = 1 + X_1 \\ & = X_2 \end{align} }[/math]

Thus, [math]\displaystyle{ X_2 }[/math] is a Kaprekar number.

b = m²k + m + 1 and p = mn + 1

Let [math]\displaystyle{ m }[/math], [math]\displaystyle{ k }[/math], and [math]\displaystyle{ n }[/math] be natural numbers, the number base [math]\displaystyle{ b = m^2k + m + 1 }[/math], and the power [math]\displaystyle{ p = mn + 1 }[/math]. Then:

• [math]\displaystyle{ X_1 = \frac{b^p - 1}{m} = (mk + 1) \sum_{i = 0}^{p - 1} b^i }[/math] is a Kaprekar number.
• [math]\displaystyle{ X_2 = \frac{b^p + m - 1}{m} = X_1 + 1 }[/math] is a Kaprekar number.

b = m²k + m + 1 and p = mn + m − 1

Let [math]\displaystyle{ m }[/math], [math]\displaystyle{ k }[/math], and [math]\displaystyle{ n }[/math] be natural numbers, the number base [math]\displaystyle{ b = m^2k + m + 1 }[/math], and the power [math]\displaystyle{ p = mn + m - 1 }[/math]. Then:

• [math]\displaystyle{ X_1 = \frac{m(b^p - 1)}{4} = (m - 1)(mk + 1) \sum_{i = 0}^{p - 1} b^i }[/math] is a Kaprekar number.
• [math]\displaystyle{ X_2 = \frac{mb^p + 1}{4} = X_3 + 1 }[/math] is a Kaprekar number.

b = m²k + m² − m + 1 and p = mn + 1

Let [math]\displaystyle{ m }[/math], [math]\displaystyle{ k }[/math], and [math]\displaystyle{ n }[/math] be natural numbers, the number base [math]\displaystyle{ b = m^2k + m^2 - m + 1 }[/math], and the power [math]\displaystyle{ p = mn + m - 1 }[/math]. Then:

• [math]\displaystyle{ X_1 = \frac{(m - 1)(b^p - 1)}{m} = (m - 1)(mk + 1) \sum_{i = 0}^{p - 1} b^i }[/math] is a Kaprekar number.
• [math]\displaystyle{ X_2 = \frac{(m - 1)b^p + 1}{m} = X_1 + 1 }[/math] is a Kaprekar number.

b = m²k + m² − m + 1 and p = mn + m − 1

Let [math]\displaystyle{ m }[/math], [math]\displaystyle{ k }[/math], and [math]\displaystyle{ n }[/math] be natural numbers, the number base [math]\displaystyle{ b = m^2k + m^2 - m + 1 }[/math], and the power [math]\displaystyle{ p = mn + m - 1 }[/math]. Then:

• [math]\displaystyle{ X_1 = \frac{b^p - 1}{m} = (mk + 1) \sum_{i = 0}^{p - 1} b^i }[/math] is a Kaprekar number.
• [math]\displaystyle{ X_2 = \frac{b^p + m - 1}{m} = X_3 + 1 }[/math] is a Kaprekar number.

Kaprekar numbers and cycles of [math]\displaystyle{ F_{p, b} }[/math] for specific [math]\displaystyle{ p }[/math], [math]\displaystyle{ b }[/math]

All numbers are in base [math]\displaystyle{ b }[/math].

Base [math]\displaystyle{ b }[/math]	Power [math]\displaystyle{ p }[/math]	Nontrivial Kaprekar numbers [math]\displaystyle{ n \neq 0 }[/math], [math]\displaystyle{ n \neq 1 }[/math]	Cycles
2	1	10	[math]\displaystyle{ \varnothing }[/math]
3	1	2, 10	[math]\displaystyle{ \varnothing }[/math]
4	1	3, 10	[math]\displaystyle{ \varnothing }[/math]
5	1	4, 5, 10	[math]\displaystyle{ \varnothing }[/math]
6	1	5, 6, 10	[math]\displaystyle{ \varnothing }[/math]
7	1	3, 4, 6, 10	[math]\displaystyle{ \varnothing }[/math]
8	1	7, 10	2 → 4 → 2
9	1	8, 10	[math]\displaystyle{ \varnothing }[/math]
10	1	9, 10	[math]\displaystyle{ \varnothing }[/math]
11	1	5, 6, A, 10	[math]\displaystyle{ \varnothing }[/math]
12	1	B, 10	[math]\displaystyle{ \varnothing }[/math]
13	1	4, 9, C, 10	[math]\displaystyle{ \varnothing }[/math]
14	1	D, 10	[math]\displaystyle{ \varnothing }[/math]
15	1	7, 8, E, 10	2 → 4 → 2 9 → B → 9
16	1	6, A, F, 10	[math]\displaystyle{ \varnothing }[/math]
2	2	11	[math]\displaystyle{ \varnothing }[/math]
3	2	22, 100	[math]\displaystyle{ \varnothing }[/math]
4	2	12, 22, 33, 100	[math]\displaystyle{ \varnothing }[/math]
5	2	14, 31, 44, 100	[math]\displaystyle{ \varnothing }[/math]
6	2	23, 33, 55, 100	15 → 24 → 15 41 → 50 → 41
7	2	22, 45, 66, 100	[math]\displaystyle{ \varnothing }[/math]
8	2	34, 44, 77, 100	4 → 20 → 4 11 → 22 → 11 45 → 56 → 45
2	3	111, 1000	10 → 100 → 10
3	3	111, 112, 222, 1000	10 → 100 → 10
2	4	110, 1010, 1111, 10000	[math]\displaystyle{ \varnothing }[/math]
3	4	121, 2102, 2222, 10000	[math]\displaystyle{ \varnothing }[/math]
2	5	11111, 100000	10 → 100 → 10000 → 1000 → 10 111 → 10010 → 1110 → 1010 → 111
3	5	11111, 22222, 100000	10 → 100 → 10000 → 1000 → 10
2	6	11100, 100100, 111111, 1000000	100 → 10000 → 100 1001 → 10010 → 1001 100101 → 101110 → 100101
3	6	10220, 20021, 101010, 121220, 202202, 212010, 222222, 1000000	100 → 10000 → 100 122012 → 201212 → 122012
2	7	1111111, 10000000	10 → 100 → 10000 → 10 1000 → 1000000 → 100000 → 1000 100110 → 101111 → 110010 → 1010111 → 1001100 → 111101 → 100110
3	7	1111111, 1111112, 2222222, 10000000	10 → 100 → 10000 → 10 1000 → 1000000 → 100000 → 1000 1111121 → 1111211 → 1121111 → 1111121
2	8	1010101, 1111000, 10001000, 10101011, 11001101, 11111111, 100000000	[math]\displaystyle{ \varnothing }[/math]
3	8	2012021, 10121020, 12101210, 21121001, 20210202, 22222222, 100000000	[math]\displaystyle{ \varnothing }[/math]
2	9	10010011, 101101101, 111111111, 1000000000	10 → 100 → 10000 → 100000000 → 10000000 → 100000 → 10 1000 → 1000000 → 1000 10011010 → 11010010 → 10011010

Extension to negative integers

Kaprekar numbers can be extended to the negative integers by use of a signed-digit representation to represent each integer.

See also

• Arithmetic dynamics
• Automorphic number
• Dudeney number
• Factorion
• Happy number
• Kaprekar's constant
• Meertens number
• Narcissistic number
• Perfect digit-to-digit invariant
• Perfect digital invariant
• Sum-product number

References

• D. R. Kaprekar (1980–1981). "On Kaprekar numbers". Journal of Recreational Mathematics 13: 81–82. 
• M. Charosh (1981–1982). "Some Applications of Casting Out 999...'s". Journal of Recreational Mathematics 14: 111–118. 
• Iannucci, Douglas E. (2000). "The Kaprekar Numbers". Journal of Integer Sequences 3: 00.1.2. Bibcode: 2000JIntS...3...12I. https://cs.uwaterloo.ca/journals/JIS/VOL3/iann2a.html.