C Program to Print Armstrong Numbers Between 1 to 1000

An Armstrong number is a number equal to the sum of its digits, each raised to the power of the total number of digits. For example, 153 and 370 are Armstrong numbers.

• The program checks each number from 1 to 1000.
• A number is printed if it satisfies the Armstrong number condition.

Examples

For 53:

153 = 1³ + 5³ + 3³
= 1 + 125 + 27
= 153

For 370:

370 = 3³ + 7³ + 0³
= 27 + 343 + 0
= 370

Approach

The program checks every number from 1 to 1000 using the following steps:

• Store the current number in a temporary variable.
• Count the number of digits in the number.
• Extract each digit and raise it to the power of the number of digits.
• Add these values to get the sum.
• If the sum is equal to the original number, print the number.

#include <stdio.h>
#include <math.h>

int main()
{
    int i, num, temp, digit, digits, sum;

    printf("Armstrong numbers between 1 and 1000 are:\n");

    for (i = 1; i <= 1000; i++) {
        num = i;
        temp = num;
        digits = 0;
        sum = 0;

        // Count the number of digits
        while (temp != 0) {
            digits++;
            temp /= 10;
        }

        temp = num;

        // Calculate the sum of powers of digits
        while (temp != 0) {
            digit = temp % 10;
            sum += pow(digit, digits);
            temp /= 10;
        }

        // Check whether the number is Armstrong
        if (sum == num) {
            printf("%d ", num);
        }
    }

    return 0;
}

Armstrong numbers between 1 and 1000 are:
1 2 3 4 5 6 7 8 9 153 370 371 407 

Explanation

• The for loop checks numbers from 1 to 1000.
• The first while loop counts the digits, while the second extracts each digit and adds its power to sum.
• pow() calculates the required power, and if sum equals the original number, it is printed.