C Program for Armstrong Numbers

We are given a task where we have to check a number n entered by the user, whether it’s Armstrong or not.

An Armstrong number is when the sum of all the digits power by the number of digits or we can say order of digits n, is same as the digit.

So below is a simple representation how to find the Armstrong number −

Formula −

wxyz…. = pow(w, n) +pow(x, n) + pow(y, n) + pow(z, n) + …..


Step 1-> Declare a function to find the value after power operation on the number
   int power(int a, int b)
      Loop while b>0
         Assign power =power * a
         Decrement b by 1
      End loop
      Return power
Step 2-> Declare a function to count the order of a number
   int count(int n)
      Declare and set i as 0
      Loop while n!=0
         Increment i by 1
         Divide n/10 and store back in n
      End loop
   Return i
Step 3-> Declare a function to check number is prime or not
   int armstrong(int n)
      Declare x and call function count(n) and assign the result to x
      Declare rem = 0 and m=0 set with zero
      Loop While n
         Set rem = n %10
         Set m = m + power(rem, x)
         Set n as n/ 10
      End Loop
   Return m;
Step 4-> Declare main
   int main(int argc, char const *argv[])
      Declare and set n = 1634
      Call function Armstrong and check if the value is equal
         Print “it is armstrong number
      End if
         Print number isn't an armstrong number


#include <stdio.h>
int power(int a, int b){
   int power =1;
      power *= a;
   return power;
int count(int n){
   int i=0;
      n = n/10;
   return i;
int armstrong(int n){
   int x = count(n);
   int rem = 0, m=0;
      rem = n %10;
      m += power(rem, x);
      n /= 10;
   return m;
int main(int argc, char const *argv[]){
   int n = 1634;
   if(n == armstrong(n)){
      printf("%d is an armstrong number 
",n);    }    else       printf("%d isn't an armstrong number
",n);    return 0; }


1634 is an armstrong number

Updated on: 23-Sep-2019


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