C Program to Check Armstrong Number

C Programming Language / Loop control in C Language

561


Program:

#include <stdio.h>
int main() {
    int num, originalNum, remainder, result = 0;
    printf("Enter a three-digit integer: ");
    scanf("%d", &num);
    originalNum = num;

    while (originalNum != 0) {
       // remainder contains the last digit
        remainder = originalNum % 10;
        
       result += remainder * remainder * remainder;
        
       // removing last digit from the orignal number
       originalNum /= 10;
    }

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

    return 0;
}

Output:

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

Explanation:

Check Armstrong Number of n digits



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

int main() {
   int num, originalNum, remainder, n = 0;
   float result = 0.0;

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

   originalNum = num;

   // store the number of digits of num in n
   for (originalNum = num; originalNum != 0; ++n) {
       originalNum /= 10;
   }

   for (originalNum = num; originalNum != 0; originalNum /= 10) {
       remainder = originalNum % 10;

      // store the sum of the power of individual digits in result
      result += pow(remainder, n);
   }

   // if num is equal to result, the number is an Armstrong number
   if ((int)result == num)
    printf("%d is an Armstrong number.", num);
   else
    printf("%d is not an Armstrong number.", num);
   return 0;
}


Output


 Enter an integer: 1634
1634 is an Armstrong number.

In this program, the number of digits of an integer is calculated first and stored in n. And, the pow() function is used to compute the power of individual digits in each iteration of the second for loop.


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