Program : Find if the given number is Armstrong
A number is said to be armstrong if the sum of cube of its digits is equal to the number itself.
e.g.
371 is an Armstrong number, since   3^3 + 7^3 + 1^3 = 371
341 is not Armstrong number, since  3^3 + 4^3 + 1^3 = 92
`import java.util.Scanner;class Armstrong{  public static void main(String args[]){     Armstrong obj = new Armstrong();     System.out.println("Enter a number to check Armstrong");      Scanner sc = new Scanner(System.in);      obj.isArmstrong(sc.nextInt());  }   public void isArmstrong(int n){     int sum = 0,number = n;     while(n!=0){        sum += Math.pow(n%10,3);       n = n/10;     }     if(sum == number)       System.out.println("Number "+number+" is Armstrong");     else       System.out.println("Number "+number+" is not Armstrong");       }}`
`Enter a number to check Armstrong371Number 371 is Armstrong`

Here we run the while loop while(n!=0) till the value of n does not becomes zero. In the loop we find the last digit of the number by n%2 and add the cube of the digit to the sum. At last if the sum is equal to the sum of cubes of digit we say the number is armstrong.
`sum    = 0    digit  = 1 number = 371sum    = 1    digit  = 7 number = 37sum    = 344  digit  = 3 number = 3sum    = 371  digit  = 0 number = 0`