Kaprekar Number
Last Updated :
11 Sep, 2023
A Kaprekar number is a number whose square when divided into two parts and such that sum of parts is equal to the original number and none of the parts has value 0. (Source : Wiki)
Given a number, the task is to check if it is Kaprekar number or not.
Examples:
Input : n = 45
Output : Yes
Explanation : 452 = 2025 and 20 + 25 is 45
Input : n = 13
Output : No
Explanation : 132 = 169. Neither 16 + 9 nor 1 + 69 is equal to 13
Input : n = 297
Output : Yes
Explanation: 2972 = 88209 and 88 + 209 is 297
Input : n = 10
Output : No
Explanation: 102 = 100. It is not a Kaprekar number even if
sum of 100 + 0 is 100. This is because of the condition that
none of the parts should have value 0.
- Find square of n and count number of digits in square.
- Split square at different positions and see if sum of two parts in any split becomes equal to n.
Below is implementation of the idea.
C++
#include<bits/stdc++.h>
using namespace std;
bool iskaprekar( int n)
{
if (n == 1)
return true ;
int sq_n = n * n;
int count_digits = 0;
while (sq_n)
{
count_digits++;
sq_n /= 10;
}
sq_n = n*n;
for ( int r_digits=1; r_digits<count_digits; r_digits++)
{
int eq_parts = pow (10, r_digits);
if (eq_parts == n)
continue ;
int sum = sq_n/eq_parts + sq_n % eq_parts;
if (sum == n)
return true ;
}
return false ;
}
int main()
{
cout << "Printing first few Kaprekar Numbers"
" using iskaprekar()\n" ;
for ( int i=1; i<10000; i++)
if (iskaprekar(i))
cout << i << " " ;
return 0;
}
|
Java
class GFG
{
static boolean iskaprekar( int n)
{
if (n == 1 )
return true ;
int sq_n = n * n;
int count_digits = 0 ;
while (sq_n != 0 )
{
count_digits++;
sq_n /= 10 ;
}
sq_n = n*n;
for ( int r_digits= 1 ; r_digits<count_digits; r_digits++)
{
int eq_parts = ( int ) Math.pow( 10 , r_digits);
if (eq_parts == n)
continue ;
int sum = sq_n/eq_parts + sq_n % eq_parts;
if (sum == n)
return true ;
}
return false ;
}
public static void main (String[] args)
{
System.out.println( "Printing first few Kaprekar Numbers" +
" using iskaprekar()" );
for ( int i= 1 ; i< 10000 ; i++)
if (iskaprekar(i))
System.out.print(i + " " );
}
}
|
Python3
import math
def iskaprekar( n):
if n = = 1 :
return True
sq_n = n * n
count_digits = 1
while not sq_n = = 0 :
count_digits = count_digits + 1
sq_n = sq_n / / 10
sq_n = n * n
r_digits = 0
while r_digits< count_digits :
r_digits = r_digits + 1
eq_parts = ( int ) (math. pow ( 10 , r_digits))
if eq_parts = = n :
continue
sum = sq_n / / eq_parts + sq_n % eq_parts
if sum = = n :
return True
return False
i = 1
while i< 10000 :
if (iskaprekar(i)) :
print (i,end = " " )
i = i + 1
|
C#
using System;
class GFG {
static bool iskaprekar( int n)
{
if (n == 1)
return true ;
int sq_n = n * n;
int count_digits = 0;
while (sq_n != 0) {
count_digits++;
sq_n /= 10;
}
sq_n = n * n;
for ( int r_digits = 1; r_digits < count_digits;
r_digits++)
{
int eq_parts = ( int )Math.Pow(10, r_digits);
if (eq_parts == n)
continue ;
int sum = sq_n / eq_parts + sq_n % eq_parts;
if (sum == n)
return true ;
}
return false ;
}
public static void Main()
{
Console.WriteLine( "Printing first few "
+ "Kaprekar Numbers using iskaprekar()" );
for ( int i = 1; i < 10000; i++)
if (iskaprekar(i))
Console.Write(i + " " );
}
}
|
PHP
<?php
function iskaprekar( $n )
{
if ( $n == 1)
return true;
$sq_n = $n * $n ;
$count_digits = 0;
while ( $sq_n )
{
$count_digits ++;
$sq_n = (int)( $sq_n / 10);
}
$sq_n1 = $n * $n ;
for ( $r_digits = 1;
$r_digits < $count_digits ;
$r_digits ++)
{
$eq_parts = pow(10, $r_digits );
if ( $eq_parts == $n )
continue ;
$sum = (int)( $sq_n1 / $eq_parts ) +
$sq_n1 % $eq_parts ;
if ( $sum == $n )
return true;
}
return false;
}
echo "Printing first few Kaprekar " .
"Numbers using iskaprekar()\n" ;
for ( $i = 1; $i < 10000; $i ++)
if (iskaprekar( $i ))
echo $i . " " ;
?>
|
Javascript
<script>
function iskaprekar(n)
{
if (n == 1)
return true ;
let sq_n = n * n;
let count_digits = 0;
while (sq_n)
{
count_digits++;
sq_n = parseInt(sq_n / 10);
}
let sq_n1 = n * n;
for (let r_digits = 1;
r_digits < count_digits;
r_digits++)
{
let eq_parts = Math.pow(10, r_digits);
if (eq_parts == n)
continue ;
let sum = parseInt((sq_n1 / eq_parts) +
sq_n1 % eq_parts);
if (sum == n)
return true ;
}
return false ;
}
document.write( "Printing first few Kaprekar " +
"Numbers using iskaprekar()<br>" );
for (let i = 1; i < 10000; i++)
if (iskaprekar(i))
document.write(i + " " );
</script>
|
Output:
Printing first few Kaprekar Numbers using iskaprekar()
1 9 45 55 99 297 703 999 2223 2728 4879 4950 5050 5292 7272 7777 9999
Time Complexity: O(log n)
Auxiliary Space: O(1)
Reference :
https://en.wikipedia.org/wiki/Kaprekar_number
Related Article:
Kaprekar Constant
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