Java program to check if a number is Armstrong

Theory

Let a₁a₂a₃…a_n a number of n digits, where a₁, a₂, a₃,…a_n are the digits of the number. The number will be called if

a₁ⁿ+a₂ⁿ+a₃ⁿ+a₄ⁿ+…….+a_nⁿ = a₁a₂a₃…a_n (the number itself)

There is only 1 digit in the number from 0 to 9. Now notice that

0¹ = 0; 1¹ = 1; 2¹ = 2; 3¹ = 3; 4¹ = 4; 5¹ = 5; 6¹ = 6; 7¹ = 7; 8¹ = 8; 9¹ = 9;

So these are all Armstrong numbers. But

1²+0² ≠ 10; 1²+1² ≠ 11; 1²+2² ≠ 12

So these three numbers and many more are not Armstrong. Here are a few more examples of Armstrong numbers:

1³+5³+3³ = 153; 3³+7³+0³ = 370; 3³+7³+1³ = 371; 4³+0³+7³ = 407

So 153, 370, 371 and 407 are also Armstrong numbers.

Java Code

package com.physicsinfo;

import java.util.Scanner;

public class Main {
    public static void main(String args[]){
        System.out.print("Enter the number you want to check for being Armstrong: ");
        Scanner sc = new Scanner(System.in);
        String str = sc.next();
        int n = str.length();
        char[] chr = new char[n];
        int[] num = new int[n];
        for (int i=0;i<n; i++){
            chr[i] = str.charAt(i);
            num[i] = Character.getNumericValue(chr[i]);
        }
        double sum = 0;
        for(int j=0;j<n;j++){
            sum = sum +Math.pow(num[j],n);
        }
        if (sum == Double.parseDouble(str)){
            System.out.println("It's a Armstrong number.");
        }
        else{
            System.out.println("It's not a Armstrong number.");
        }
    }
}