Kaprekar_number

Kaprekar number

Base-dependent property of integers

In mathematics, a natural number in a given number base is a $p$ -Kaprekar number if the representation of its square in that base can be split into two parts, where the second part has $p$ 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 $n$ be a natural number. We define the Kaprekar function for base $b>1$ and power $p>0$ $F_{p,b}:\mathbb {N} \rightarrow \mathbb {N}$ to be the following:

F_{p,b}(n)=\alpha +\beta

,

where $\beta =n^{2}{\bmod {b}}^{p}$ and

\alpha ={\frac {n^{2}-\beta }{b^{p}}}

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

The earlier example of 45 satisfies this definition with $b=10$ and $p=2$ , because

\beta =n^{2}{\bmod {b}}^{p}=45^{2}{\bmod {1}}0^{2}=25

\alpha ={\frac {n^{2}-\beta }{b^{p}}}={\frac {45^{2}-25}{10^{2}}}=20

F_{2,10}(45)=\alpha +\beta =20+25=45

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

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

There are only a finite number of $p$ -Kaprekar numbers and cycles for a given base $b$ , because if $n=b^{p}+m$ , where $m>0$ then

{\begin{aligned}n^{2}&=(b^{p}+m)^{2}\\&=b^{2p}+2mb^{p}+m^{2}\\&=(b^{p}+2m)b^{p}+m^{2}\\\end{aligned}}

and $\beta =m^{2}$ , $\alpha =b^{p}+2m$ , and $F_{p,b}(n)=b^{p}+2m+m^{2}=n+(m^{2}+m)>n$ . Only when $n\leq b^{p}$ do Kaprekar numbers and cycles exist.

If $d$ is any divisor of $p$ , then $n$ is also a $p$ -Kaprekar number for base $b^{p}$ .

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

Set-theoretic definition and unitary divisors

We can define the set $K(N)$ for a given integer $N$ as the set of integers $X$ for which there exist natural numbers $A$ and $B$ satisfying the Diophantine equation^[1]

X^{2}=AN+B

, where

0\leq B<N

X=A+B

An $n$ -Kaprekar number for base $b$ is then one which lies in the set $K(b^{n})$ .

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

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

Kaprekar numbers for $F_{p,b}$

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

Let $k$ and $n$ be natural numbers, the number base $b=4k+3=2(2k+1)+1$ , and $p=2n+1$ . Then:

$X_{1}={\frac {b^{p}-1}{2}}=(2k+1)\sum _{i=0}^{p-1}b^{i}$ is a Kaprekar number.

Proof

Let

${\begin{aligned}X_{1}&={\frac {b^{p}-1}{2}}\\&={\frac {b-1}{2}}\sum _{i=0}^{p-1}b^{i}\\&={\frac {4k+3-1}{2}}\sum _{i=0}^{2n+1-1}b^{i}\\&=(2k+1)\sum _{i=0}^{2n}b^{i}\end{aligned}}$

Then,

${\begin{aligned}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{aligned}}$

The two numbers $\alpha$ and $\beta$ are

\beta =X_{1}^{2}{\bmod {b}}^{p}=k+1+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}

\alpha ={\frac {X_{1}^{2}-\beta }{b^{p}}}=k+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}

and their sum is

${\begin{aligned}\alpha +\beta &=\left(k+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)+\left(k+1+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)\\&=2k+1+\sum _{i=1}^{n}((2k)b+2(3k+2))b^{2i-1}\\&=2k+1+\sum _{i=1}^{n}((2k)b+(6k+4))b^{2i-1}\\&=2k+1+\sum _{i=1}^{n}((2k)b+(4k+3))b^{2i-1}+(2k+1)b^{2i-1}\\&=2k+1+\sum _{i=1}^{n}((2k+1)b)b^{2i-1}+(2k+1)b^{2i-1}\\&=2k+1+\sum _{i=1}^{n}(2k+1)b^{2i}+(2k+1)b^{2i-1}\\&=2k+1+\sum _{i=1}^{2n}(2k+1)b^{i}\\&=\sum _{i=0}^{2n}(2k+1)b^{i}\\&=(2k+1)\sum _{i=0}^{2n}b^{i}&=X_{1}\\\end{aligned}}$

Thus, $X_{1}$ is a Kaprekar number.

$X_{2}={\frac {b^{p}+1}{2}}=X_{1}+1$ is a Kaprekar number for all natural numbers $n$ .

Proof

Let

${\begin{aligned}X_{2}&={\frac {b^{2n+1}+1}{2}}\\&={\frac {b^{2n+1}-1}{2}}+1\\&=X_{1}+1\end{aligned}}$

Then,

${\begin{aligned}X_{2}^{2}&=(X_{1}+1)^{2}\\&=X_{1}^{2}+2X_{1}+1\\&=X_{1}^{2}+2X_{1}+1\\&=b^{p}\left(k+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)+\left(k+1+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)+b^{p}-1+1\\&=b^{p}\left(k+1+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)+\left(k+1+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)\end{aligned}}$

The two numbers $\alpha$ and $\beta$ are

\beta =X_{2}^{2}{\bmod {b}}^{p}=k+1+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}

\alpha ={\frac {X_{2}^{2}-\beta }{b^{p}}}=k+1+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}

and their sum is

${\begin{aligned}\alpha +\beta &=\left(k+1+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)+\left(k+1+\sum _{i=1}^{n}(kb+(3k+2))b^{2i-1}\right)\\&=2k+2+\sum _{i=1}^{n}((2k)b+2(3k+2))b^{2i-1}\\&=2k+2+\sum _{i=1}^{n}((2k)b+(6k+4))b^{2i-1}\\&=2k+2+\sum _{i=1}^{n}((2k)b+(4k+3))b^{2i-1}+(2k+1)b^{2i-1}\\&=2k+2+\sum _{i=1}^{n}((2k+1)b)b^{2i-1}+(2k+1)b^{2i-1}\\&=2k+2+\sum _{i=1}^{n}(2k+1)b^{2i}+(2k+1)b^{2i-1}\\&=2k+2+\sum _{i=1}^{2n}(2k+1)b^{i}\\&=1+\sum _{i=0}^{2n}(2k+1)b^{i}\\&=1+(2k+1)\sum _{i=0}^{2n}b^{i}\\&=1+X_{1}\\&=X_{2}\end{aligned}}$

Thus, $X_{2}$ is a Kaprekar number.

b = m²k + m + 1 and p = mn + 1

Let $m$ , $k$ , and $n$ be natural numbers, the number base $b=m^{2}k+m+1$ , and the power $p=mn+1$ . Then:

$X_{1}={\frac {b^{p}-1}{m}}=(mk+1)\sum _{i=0}^{p-1}b^{i}$ is a Kaprekar number.
$X_{2}={\frac {b^{p}+m-1}{m}}=X_{1}+1$ is a Kaprekar number.

b = m²k + m + 1 and p = mn + m − 1

Let $m$ , $k$ , and $n$ be natural numbers, the number base $b=m^{2}k+m+1$ , and the power $p=mn+m-1$ . Then:

$X_{1}={\frac {m(b^{p}-1)}{4}}=(m-1)(mk+1)\sum _{i=0}^{p-1}b^{i}$ is a Kaprekar number.
$X_{2}={\frac {mb^{p}+1}{4}}=X_{3}+1$ is a Kaprekar number.

b = m²k + m² − m + 1 and p = mn + 1

Let $m$ , $k$ , and $n$ be natural numbers, the number base $b=m^{2}k+m^{2}-m+1$ , and the power $p=mn+m-1$ . Then:

$X_{1}={\frac {(m-1)(b^{p}-1)}{m}}=(m-1)(mk+1)\sum _{i=0}^{p-1}b^{i}$ is a Kaprekar number.
$X_{2}={\frac {(m-1)b^{p}+1}{m}}=X_{1}+1$ is a Kaprekar number.

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

Let $m$ , $k$ , and $n$ be natural numbers, the number base $b=m^{2}k+m^{2}-m+1$ , and the power $p=mn+m-1$ . Then:

$X_{1}={\frac {b^{p}-1}{m}}=(mk+1)\sum _{i=0}^{p-1}b^{i}$ is a Kaprekar number.
$X_{2}={\frac {b^{p}+m-1}{m}}=X_{3}+1$ is a Kaprekar number.

Kaprekar numbers and cycles of $F_{p,b}$ for specific $p$ , $b$

All numbers are in base $b$ .

More information Base

...

Base $b$	Power $p$	Nontrivial Kaprekar numbers $n\neq 0$ , $n\neq 1$	Cycles
2	1	10	$\varnothing$
3	1	2, 10	$\varnothing$
4	1	3, 10	$\varnothing$
5	1	4, 5, 10	$\varnothing$
6	1	5, 6, 10	$\varnothing$
7	1	3, 4, 6, 10	$\varnothing$
8	1	7, 10	2 → 4 → 2
9	1	8, 10	$\varnothing$
10	1	9, 10	$\varnothing$
11	1	5, 6, A, 10	$\varnothing$
12	1	B, 10	$\varnothing$
13	1	4, 9, C, 10	$\varnothing$
14	1	D, 10	$\varnothing$
15	1	7, 8, E, 10	2 → 4 → 2 9 → B → 9
16	1	6, A, F, 10	$\varnothing$
2	2	11	$\varnothing$
3	2	22, 100	$\varnothing$
4	2	12, 22, 33, 100	$\varnothing$
5	2	14, 31, 44, 100	$\varnothing$
6	2	23, 33, 55, 100	15 → 24 → 15 41 → 50 → 41
7	2	22, 45, 66, 100	$\varnothing$
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	$\varnothing$
3	4	121, 2102, 2222, 10000	$\varnothing$
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	$\varnothing$
3	8	2012021, 10121020, 12101210, 21121001, 20210202, 22222222, 100000000	$\varnothing$
2	9	10010011, 101101101, 111111111, 1000000000	10 → 100 → 10000 → 100000000 → 10000000 → 100000 → 10 1000 → 1000000 → 1000 10011010 → 11010010 → 10011010

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This article uses material from the Wikipedia article Kaprekar_number, and is written by contributors. Text is available under a CC BY-SA 4.0 International License; additional terms may apply. Images, videos and audio are available under their respective licenses.

[iannucci-1] [1]
Iannucci (2000)

[1]

Kaprekar_number

Kaprekar number

Definition and properties

Set-theoretic definition and unitary divisors