C program to find Armstrong numbers between 1 to n

Write a C program to enter any number and print all Armstrong numbers between 1 to n. How to print Armstrong numbers between given interval using loop in C program. Logic to generate Armstrong numbers in given range in C program.

Enter lower limit: 1
Enter upper limit: 1000

Armstrong number between 1 to 1000 are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 370, 371, 407

Required knowledge

Basic C programming, For loop, Nested loop, If else

What is Armstrong number?

An Armstrong number is an n-digit number that is equal to the sum of the n^th powers of its digits. Read more about Armstrong numbers. Below are few examples of Armstrong numbers:
6 = 6¹ = 6
371 = 3³ + 7³ + 1³ = 371

Logic to generate Armstrong number from 1 to n

Below is the step by step descriptive logic to generate Armstrong numbers:

1. Read upper limit to generate Armstrong number from user. Store it in some variable say end.
2. Run a loop from 1 to end, incrementing 1 in each iteration. The loop structure should look like for(i=1; i<=end; i++). This loop will iterate through n numbers. Inside this loop I will check each number for Armstrong number.
3. Inside the loop print the current number if it is Armstrong number. Read the below post for details about checking Armstrong number.
   • Program to check Armstrong number.

Program to generate Armstrong numbers from 1 to n

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

int main()
{
    int num, lastDigit, digits, sum, i, end;

    /*
     * Read upper limit from user
     */
    printf("Enter upper limit: ");
    scanf("%d", &end);

    printf("Armstrong number between 1 to %d are: \n", end);

    for(i=1; i<=end; i++)
    {
        sum = 0;

        // Copy the value of num for processing
        num = i;

        /*
         * Find total digits in num
         */
        digits = (int) log10(num) + 1;
        /*
         * Calculate sum of power of digits
         */
        while(num > 0)
        {
            // Extract the last digit
            lastDigit = num % 10;

            // Find sum
            sum = sum + pow(lastDigit, digits);

            // Remove the last digit
            num = num / 10;
        }

        // Check for Armstrong number
        if(i == sum)
        {
            printf("%d, ", i);
        }

    }

    return 0;
}

So whats next now? Once you are done with generating Armstrong numbers from 1 to n. You can easily modify the logic to work it for given ranges. Below program generates Armstrong numbers in a given range.

Program to find Armstrong numbers in given range

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

int main()
{
    int num, lastDigit, digits, sum, i;
    int start, end;

    /*
     * Read lower and upper limit from user
     */
    printf("Enter lower limit: ");
    scanf("%d", &start);
    printf("Enter upper limit: ");
    scanf("%d", &end);

    printf("Armstrong number between %d to %d are: \n", start, end);

    for(i=start; i<=end; i++)
    {
        sum = 0;

        // Copy the value of num for processing
        num = i;

        /*
         * Find total digits in num
         */
        digits = (int) log10(num) + 1;
        /*
         * Calculate sum of power of digits
         */
        while(num > 0)
        {
            // Extract the last digit
            lastDigit = num % 10;

            // Find sum
            sum = sum + pow(lastDigit, digits);

            // Remove the last digit
            num = num / 10;
        }

        // Check for Armstrong number
        if(i == sum)
        {
            printf("%d, ", i);
        }

    }

    return 0;
}

Enter lower limit: 1
Enter upper limit: 10000
Armstrong number between 1 to 10000 are:
1, 2, 3, 4, 5, 6, 7, 8, 9, 370, 371, 407, 1634, 8208, 9474,

Happy coding ;)