A number can be called as an Armstrong number only if the sum of cubes of its digits is equal to the number itself.

Here in this c program, we are going to check whether the input of the number is Armstrong or not.

Let’s take a case where an Armstrong number consists of 3 digits. so let us try to understand with an example considering the number to be 153

**example:**

153 = (1*1*1) + (5*5*5) + (3*3*3) That is: (1*1*1) = 1 (5*5*5) = 125 (3*3*3) = 27 So: 1 + 125 + 27 = 153

hence we can Consider 153 as an Armstrong number.

Now let us move to our check out our program to check Armstrong Number in C.

## 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: 153 371 is an Armstrong number.

## Example #2: Check Armstrong Number of n digits

#include<stdio.h> int main() { int n,r,sum=0,temp; printf("enter the number="); scanf("%d",&n); temp=n; while(n>0) { r=n%10; sum=sum+(r*r*r); n=n/10; } if(temp==sum) printf("armstrong number."); else printf("not Armstrong number"); return 0; }

**Output**

Enter an integer: 1221 Armstrong number.

In short, this c program tends to calculate the number of digits of an integer and stores in a variable named n. And then uses the pow() function to find the power of individual digits using while loop.