Kaprekar Number

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.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 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

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

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

C#

 // C# program to check if a number is // Kaprekar number or not using System;    class GFG {            // Returns true if n is a Kaprekar      // number, else false     static 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 != 0) {             count_digits++;             sq_n /= 10;         }            // Recompute square as it was changed         sq_n = n * n;                     // 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()     {                    Console.WriteLine("Printing first few "          + "Kaprekar Numbers using iskaprekar()");            for (int i = 1; i < 10000; i++)             if (iskaprekar(i))                 Console.Write(i + " ");     } }    // This code is contributed by vt_m.

PHP



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

Related Article:
Kaprekar Constant

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