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

C For Loop: Exercise-59 with Solution

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

Pictorial Presentation:

Sample Solution:

C Code:

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

 
int main()
{
    int n1, onum, r, result = 0, n = 0 ;
    printf("\n\n Check whether an n digits number is armstrong or not :\n");
	printf("-----------------------------------------------------------\n"); 	

    printf(" Input  an integer : ");
    scanf("%d", &n1);

     onum = n1;
    
    while (onum != 0)
    {
        onum /= 10;
        ++n;
    }

    onum = n1;

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

    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:

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