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Print all Armstrong numbers between 1 and N

Introduction

An Armstrong number (also called a narcissistic number) is a number that is equal to the sum of its digits raised to the power of the number of digits.

Examples:

  • 153 = 1³ + 5³ + 3³ = 153 Armstrong Number)
  • 370 = 3³ + 7³ + 0³ = 370 Armstrong Number)
  • 123 = 1³ + 2³ + 3³ = 36 (Not Armstrong Number)

In this program, the user enters a positive integer N, and the program prints all Armstrong numbers from 1 to N using nested for, while and do..while loops.

 

C Program: Print all Armstrong numbers between 1 and N

Method 1: Using for loop

C 
#include <stdio.h>


#include <math.h>


 


int main() {


    int n, num, original, remainder, digits;


    double sum;


 


    // Input


    printf("Enter the value of N: ");


    scanf("%d", &n);


 


    // Validate input


    if (n <= 0) {


        printf("Invalid input! Please enter a positive number.\n");


        return 0;


    }


 


    printf("Armstrong numbers between 1 and %d are:\n", n);


 


    // Loop through numbers from 1 to n


    for (num = 1; num <= n; num++) {


        original = num;


        sum = 0;


        digits = 0;


 


        // Count number of digits


        for (int temp = num; temp != 0; temp /= 10)


            digits++;


 


        // Calculate sum of powers of digits


        for (int temp = num; temp != 0; temp /= 10) {


            remainder = temp % 10;


            sum += pow(remainder, digits);


        }


 


        // Check Armstrong condition


        if ((int)sum == original)


            printf("%d ", original);


    }


 


    printf("\n");


    return 0;


}

Output

 
OUTPUT 1 :
Enter the value of N: 500
Armstrong numbers between 1 and 500 are:
1 2 3 4 5 6 7 8 9 153 370 371 407

OUTPUT 2 :
Enter the value of N: 200
Armstrong numbers between 1 and 200 are:
1 2 3 4 5 6 7 8 9 153

Explanation

  1. The user enters a positive integer n.
  2. Outer loop iterates each number from 1 to n.
  3. For each number:
    • Count the number of digits.
    • Compute the sum of each digit raised to the power of the number of digits.
  4. If the sum equals the number → it is an Armstrong number and printed.

 

C Program: Print all Armstrong numbers between 1 and N

Method 2: Using while loop

C 
#include <stdio.h>


#include <math.h>


 


int main() {


    int n, num, original, remainder, digits, temp;


    double sum;


 


    // Input


    printf("Enter the value of N: ");


    scanf("%d", &n);


 


    // Validate input


    if (n <= 0) {


        printf("Invalid input! Please enter a positive number.\n");


        return 0;


    }


 


    printf("Armstrong numbers between 1 and %d are:\n", n);


 


    // Loop through numbers from 1 to n


    num = 1;


    while (num <= n) {


        original = num;


        sum = 0;


        digits = 0;


        temp = num;


 


        // Count number of digits


        while (temp != 0) {


            digits++;


            temp /= 10;


        }


 


        // Calculate sum of powers of digits


        temp = num;


        while (temp != 0) {


            remainder = temp % 10;


            sum += pow(remainder, digits);


            temp /= 10;


        }


 


        // Check Armstrong condition


        if ((int)sum == original)


            printf("%d ", original);


 


        num++;


    }


 


    printf("\n");


    return 0;


}

Output

 
OUTPUT 1 :
Enter the value of N: 500
Armstrong numbers between 1 and 500 are:
1 2 3 4 5 6 7 8 9 153 370 371 407

OUTPUT 2 :
Enter the value of N: 200
Armstrong numbers between 1 and 200 are:
1 2 3 4 5 6 7 8 9 153

Explanation

  1. User inputs a positive integer n.
  2. Outer while loop iterates through numbers from 1 to n.
  3. For each number:
    • Count digits using a while
    • Compute sum of each digit raised to the number of digits using another while
  4. If sum equals the number → it’s an Armstrong number and printed.

 

C Program: Print all Armstrong numbers between 1 and N

Method 3: Using do..while loop

C 
#include <stdio.h>


#include <math.h>


 


int main() {


    int n, num = 1, original, remainder, digits, temp;


    double sum;


 


    // Input


    printf("Enter the value of N: ");


    scanf("%d", &n);


 


    // Validate input


    if (n <= 0) {


        printf("Invalid input! Please enter a positive number.\n");


        return 0;


    }


 


    printf("Armstrong numbers between 1 and %d are:\n", n);


 


    // Loop through numbers from 1 to n


    do {


        original = num;


        sum = 0;


        digits = 0;


        temp = num;


 


        // Count number of digits


        if (temp == 0)


            digits = 1; // special case for 0


        else


            do {


                digits++;


                temp /= 10;


            } while (temp != 0);


 


        // Calculate sum of powers of digits


        temp = num;


        do {


            remainder = temp % 10;


            sum += pow(remainder, digits);


            temp /= 10;


        } while (temp != 0);


 


        // Check Armstrong condition


        if ((int)sum == original)


            printf("%d ", original);


 


        num++;


    } while (num <= n);


 


    printf("\n");


    return 0;


}

Output

 
OUTPUT 1 :
Enter the value of N: 500
Armstrong numbers between 1 and 500 are:
1 2 3 4 5 6 7 8 9 153 370 371 407

OUTPUT 2 :
Enter the value of N: 200
Armstrong numbers between 1 and 200 are:
1 2 3 4 5 6 7 8 9 153

Explanation

  1. User inputs a positive integer n.
  2. Outer do..while loop iterates num from 1 to n.
  3. For each number:
    • Count digits using a do..while loop.
    • Compute sum of each digit raised to the number of digits using another do..while loop.
  4. If sum equals the number → it’s an Armstrong number and printed.

 

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