C Exercises: Check whether an n digits number is Armstrong or not

C For Loop: Exercise-59 with Solution

Write a C program to check the Armstrong number of n digits.

Visual Presentation:

Sample Solution:

C Code:

#include <stdio.h>   // Include the standard input/output header file. 

#include <math.h>     // Include the math header file.
int main()
{
    // Variable declarations
int n1, onum, r, result = 0, n = 0 ;
    // Prompting user for input
printf("\n\n Check whether an n digits number is armstrong or not :\n");
printf("-----------------------------------------------------------\n"); 	
printf(" Input an integer : ");
scanf("%d", &n1);
    // Store the original number for later comparison
onum = n1;
    // Count the number of digits in the input number
while (onum != 0)
    {
onum /= 10;
        ++n;
    }
    // Reset onum to the original number for further processing
onum = n1;
    // Calculate the sum of cubes of individual digits raised to the power of 'n'
while (onum != 0)
    {
        r = onum % 10;
result += pow(r, n);
onum /= 10;
    }
    // Check if the result is equal to the original number
if(result == n1)
printf(" %d is an Armstrong number.\n\n", n1);
else
printf(" %d is not an Armstrong number.\n\n", n1);
return 0;
}

Sample Output:

 Check whether an n digits number is Armstrong or not :                                                       
-----------------------------------------------------------                                                   
 Input  an integer : 1634                                                                                     
 1634 is an Armstrong number.                                                                                                           

Flowchart:

C Programming Code Editor:

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