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

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

41. Keith Number Check

Write a C++ program to check if a number is Keith or not (with explanation).

Sample Output:
Input a number : 197
1 + 9 + 7 = 17
9 + 7 + 17 = 33
7 + 17 + 33 = 57
17 + 33 + 57 = 107
33 + 57 + 107 = 197
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

// Function to count the number of digits in a number
int lenCount(int nm) {
    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
}

// Main function
int main() {
    int num1 = 0, arr1[10], num2 = 0, flg = 0, i = 0, sum = 0, kk; // 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) {
        for (i = 0; i < lenCount(num2); i++)
            sum += arr1[i]; // Calculate the sum of the digits in the array

        // If the sum matches the original number, it's a Keith number
        if (sum == num2) {
            flg = 1; // Set the flag to terminate the loop
            kk = 1; // Set kk to 1 indicating the number is a Keith number
        }

        // If the sum exceeds the original number, it's not a Keith number
        if (sum > num2) {
            flg = 1; // Set the flag to terminate the loop
            kk = 0; // Set kk to 0 indicating the number is not a Keith number
        }

        // Loop to reconstruct the array by shifting elements and adding the new sum
        for (i = 0; i < lenCount(num2); i++) {
            cout << " " << arr1[i]; // Display the array elements
            if (i != lenCount(num2) - 1) {
                arr1[i] = arr1[i + 1]; // Shift elements in the array
                cout << " + "; // Displaying '+' between elements
            } else {
                arr1[i] = sum; // Store the new sum at the last index of the array
                cout << " = " << arr1[i]; // Display the sum
            }
        }
        cout << endl; // Move to the next line
        sum = 0; // Reset the sum for the next iteration
    }

    // Displaying whether the number is a Keith number or not based on kk value
    if (kk == 1) {
        cout << " The given number is a Keith Number.\n"; // Displaying output
    }
    if (kk == 0) {
        cout << " The given number is not a Keith Number.\n"; // Displaying output
    }
}

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 : 197                                                  
 1 +  9 +  7 = 17                                                      
 9 +  7 +  17 = 33                                                     
 7 +  17 +  33 = 57                                                    
 17 +  33 +  57 = 107                                                  
 33 +  57 +  107 = 197                                                 
 The given number is a Keith Number.

Flowchart:

For more Practice: Solve these Related Problems:

Write a C++ program to determine if a number is a Keith number by generating its digit sequence iteratively.
Write a C++ program to check for Keith numbers using a vector to store and update the sequence dynamically.
Write a C++ program to verify Keith numbers with recursive functions that simulate the additive process.
Write a C++ program to test if a number is Keith by comparing the computed sequence against the original number with early termination.

C++ Code Editor:

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