C++ Exercises: Check if a number is Keith or not

Last update on April 12 2025 13:00:31 (UTC/GMT +8 hours)

Find Strong Numbers Within a Range

Write a C++ program to check if a number is Keith or not.

Sample Output:
Input a number : 742
The given number is a Keith Number.

Sample Solution:

C++ Code :

#include<bits/stdc++.h> // Including the complete set of standard C++ libraries

using namespace std; // Using the standard namespace

int lenCount(int nm) // Function to count the number of digits in a number
{
    int ctr = 0; // Counter initialized to count the digits
    while (nm > 0) // Loop to count digits
    {
        nm = nm / 10; // Remove the last digit
        ctr++; // Increment the counter for each digit removed
    }
    return ctr; // Return the count of digits
}

int main() // Main function
{
    int num1 = 0, arr1[10], num2 = 0, flg = 0, i = 0, sum = 0; // Declaration of integer variables

    cout << "\n\n Check whether a number is Keith or not: \n"; // Displaying purpose
    cout << " Sample Keith numbers: 197, 742, 1104, 1537, 2208, 2580, 3684, 4788, 7385..\n"; // Displaying purpose
    cout << " -----------------------------------------------------------------------\n"; // Displaying purpose	
    cout << " Input a number : "; // Prompting user for input
    cin >> num1; // Taking user input

    num2 = num1; // Storing the original number

    // Loop to store individual digits of the number into an array in reverse order
    for (i = lenCount(num2) - 1; i >= 0; i--)
    {
        arr1[i] = num1 % 10; // Extract the last digit
        num1 /= 10; // Remove the last digit
    }

    // Loop to check if the number is a Keith number
    while (flg == 0)
    {
        // Calculate the sum of the digits in the array
        for (i = 0; i < lenCount(num2); i++)
            sum += arr1[i];

        // If the sum matches the original number, it's a Keith number
        if (sum == num2)
        {
            cout << " The given number is a Keith Number.\n"; // Displaying output
            flg = 1; // Set the flag to terminate the loop
        }

        // If the sum exceeds the original number, it's not a Keith number
        if (sum > num2)
        {
            cout << " The given number is not a Keith number.\n"; // Displaying output
            flg = 1; // Set the flag to terminate the loop
        }

        // Reconstruct the array by shifting elements and adding the new sum
        for (i = 0; i < lenCount(num2); i++)
        {
            if (i != lenCount(num2) - 1)
                arr1[i] = arr1[i + 1]; // Shift elements in the array
            else
                arr1[i] = sum; // Store the new sum at the last index of the array
        }
        sum = 0; // Reset the sum for the next iteration
    }
}

Sample Output:

 Check whether a number is Keith or not:                               
 Sample Keith numbers: 197, 742, 1104, 1537, 2208, 2580, 3684, 4788, 7385..                                                                   
 -----------------------------------------------------------------------                                                                     
 Input a number : 742                                                  
 The given number is a Keith Number.

Flowchart:

For more Practice: Solve these Related Problems:

Write a C++ program to list all strong numbers in a given range by comparing the sum of factorials of digits.
Write a C++ program to identify strong numbers between two limits using efficient factorial calculation.
Write a C++ program to generate strong numbers within a range with inline functions for factorial computation.
Write a C++ program to display strong numbers and count them using STL algorithms for processing digit factorials.

C++ Code Editor:

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