Question

Kaprekar C++ program

a. Write a c++  code that takes any number between 0 and 9999 as an input value, and returns the number of iterations it takes that number to reach 6174. Also, it needs to indicate if the number does not converge to 6174. Run the program with three different numbers of your choosing to show that it works.

b. Write a second c++ code using the first. However, this new program will calculate how many iterations it takes each number from 0 to 9999 to converge to 6174. The output will be (1) the number of values that do not converge within 8 iterations, and (2) a table with the first column being iterations from 1 through 8, and the second column being the number of values that took that many times to converge. For example, the number of values that converge after three calculations is 2400.

Answer 1

a)

CODE

//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()

{

int n;

cout << "Enter a number between 0 and 9999: ";

cin >> n;

int count = 0;

while(n <= 6174) {

if (n == 6174) {

cout << "Number of iterations: " << count;

return 0;

}

n = iskaprekar(n);

count ++;

}

  return 0;

}

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