Program to check if the given number is Armstrong
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 Armstrong
371
Number 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 = 371
sum    = 1    digit  = 7 number = 37
sum    = 344  digit  = 3 number = 3
sum    = 371  digit  = 0 number = 0
```