Write a C program to check whether a given number is an Armstrong number or not

Solution:

• A positive integer is called an Armstrong number (of order n) if xyz… = x^n + y^n + z^n
• In the case of an Armstrong number of 3 digits, the sum of cubes of each digit is equal to the number itself.
• For example, 153 is an Armstrong number because 153 = 1*1*1 + 5*5*5 + 3*3*3

#include<stdio.h>
int main()    
{    
	int n,r,sum=0,temp;    
	printf("Enter n value : ");    
	scanf("%d",&n);    
	temp=n;    
	while(n>0)    
	{    
		r=n%10;    
		sum=sum+(r*r*r);    
		n=n/10;    
	}   
	n=temp; 
	if(temp==sum)    
		printf("%d is Armstrong  number \n", n);    
	else    
		printf("%d is Not an armstrong number\n", n);    
	return 0;  
}

Approach-2:

• In the first approach, we check only 3 digits number is strong number or not.
• In the following approach, we can mention any number of digits to check whether Armstrong or not.
• In the code, we write custom function to check the order of number as well as to find the power of input number.

#include <stdio.h> 
int power(int, int);
int order(int);
int main() 
{ 
int n=153, sum=0, temp, r;
int d = digits(n);
printf("Input number is : %d \n", n);
printf("Digits : %d\n", d);
temp=n;
while(n)
{
r = n%10;
sum += power(r, d);
n = n/10;
}
n=temp;
if(n==sum)
printf("Armstrong number\n");
else
printf("Not an ArmStrong number\n");
return 0;
} 
int digits(int x) 
{ 
int d = 0;
while(x)
{
d++;
x=x/10;
}
return d;
}								
int power(int x, int y) 
{ 
	int i, res=x;
	for(i=1 ; i<y ; i++)
	{
		res = res*x;
	}
    return res;
}