Kaprekar Number

3

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.

  1. Find square of n and count number of digits in square.
  2. 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/C++

//C program to check if a number is Kaprekar number or not
#include<bits/stdc++.h>
using namespace std;

// Returns true if n is a Kaprekar number, else false
bool iskaprekar(int n)
{
    if (n == 1)
       return true;

    // Count number of digits in square
    int sq_n = n * n;
    int count_digits = 0;
    while (sq_n)
    {
        count_digits++;
        sq_n /= 10;
    }

    sq_n = n*n; // Recompute square as it was changed

    // Split the square at different poitns and see if sum
    // of any pair of splitted numbers is equal to n.
    for (int r_digits=1; r_digits<count_digits; r_digits++)
    {
         int eq_parts = pow(10, r_digits);

         // To avoid numbers like 10, 100, 1000 (These are not
         // Karprekar numbers
         if (eq_parts == n)
            continue;

         // Find sum of current parts and compare with n
         int sum = sq_n/eq_parts + sq_n % eq_parts;
         if (sum == n)
           return true;
    }

    // compare with original number
    return false;
}

// Driver code
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

// Java program to check if a number is 
// Kaprekar number or not

class GFG 
{
	// Returns true if n is a Kaprekar number, else false
	static boolean iskaprekar(int n)
	{
	    if (n == 1)
	       return true;
	 
	    // Count number of digits in square
	    int sq_n = n * n;
	    int count_digits = 0;
	    while (sq_n != 0)
	    {
	        count_digits++;
	        sq_n /= 10;
	    }
	 
	    sq_n = n*n; // Recompute square as it was changed
	 
	    // Split the square at different poitns and see if sum
	    // of any pair of splitted numbers is equal to n.
	    for (int r_digits=1; r_digits<count_digits; r_digits++)
	    {
	         int eq_parts = (int) Math.pow(10, r_digits);
	 
	         // To avoid numbers like 10, 100, 1000 (These are not
	         // Karprekar numbers
	         if (eq_parts == n)
	            continue;
	 
	         // Find sum of current parts and compare with n
	         int sum = sq_n/eq_parts + sq_n % eq_parts;
	         if (sum == n)
	           return true;
	    }
	 
	    // compare with original number
	    return false;
	}
	
    // Driver method
	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 + " ");
	}
}

Python

# Python program to check if a number is Kaprekar number or not

import math

# Returns true if n is a Kaprekar number, else false
def iskaprekar( n):
    if n == 1 :
        return True
    
    #Count number of digits in square
    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  # Recompute square as it was changed
    
    # Split the square at different poitns and see if sum
    # of any pair of splitted numbers is equal to n.
    r_digits = 0
    while r_digits< count_digits :
        r_digits = r_digits + 1
        eq_parts = (int) (math.pow(10, r_digits))
        
        # To avoid numbers like 10, 100, 1000 (These are not
        # Karprekar numbers
        if eq_parts == n :
            continue
        
        # Find sum of current parts and compare with n
        
        sum = sq_n/eq_parts + sq_n % eq_parts
        if sum == n :
            return True
    
    # compare with original number
    return False
    
# Driver method
i=1
while i<10000 :
    if (iskaprekar(i)) :
        print i," ",
    i = i + 1
# code contributed by Nikita Tiwari


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 

Reference :
https://en.wikipedia.org/wiki/Kaprekar_number

Related Article:
Kaprekar Constant

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