Check Armstrong Number

Example to check whether an integer (entered by the user) is an Armstrong number or not using while loop and if...else statement.

A positive integer is called an Armstrong number of order n if

abcd... = an + bn + cn + dn + ...

In case of an Armstrong number of 3 digits, the sum of cubes of each digits is equal to the number itself. For example:

153 = 1*1*1 + 5*5*5 + 3*3*3  // 153 is an Armstrong number.

Example #1: Check Armstrong Number of three digits

#include <stdio.h>
int main()
{
    int number, originalNumber, remainder, result = 0;

    printf("Enter a three digit integer: ");
    scanf("%d", &number);

    originalNumber = number;

    while (originalNumber != 0)
    {
        remainder = originalNumber%10;
        result += remainder*remainder*remainder;
        originalNumber /= 10;
    }

    if(result == number)
        printf("%d is an Armstrong number.",number);
    else
        printf("%d is not an Armstrong number.",number);

    return 0;
}

Output

Enter a three digit integer: 371
371 is an Armstrong number.

Example #2: Check Armstrong Number of n digits

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

int main()
{
    int number, originalNumber, remainder, result = 0, n = 0 ;

    printf("Enter an integer: ");
    scanf("%d", &number);

     originalNumber = number;
    
    while (originalNumber != 0)
    {
        originalNumber /= 10;
        ++n;
    }

    originalNumber = number;

    while (originalNumber != 0)
    {
        remainder = originalNumber%10;
        result += pow(remainder, n);
        originalNumber /= 10;
    }

    if(result == number)
        printf("%d is an Armstrong number.", number);
    else
        printf("%d is not an Armstrong number.", number);

    return 0;
}

 

Output

Enter an integer: 1634
1634 is an Armstrong number.