Question

I need some help checking over my program to find the digit root. If a user enters 5635 the digit root is 1. To find the digit root of a number you add all the digits in the number 5 + 6 + 3 + 5 and you get the result of 19. Then you add the result 1 + 9 = 10. Then you add 1 + 0 until you get 1 which is your digit root.

1. Do i have the right approach or is my approach to the question totally off?
2. Why am i getting 0 as the result instead of the correct answer of 1?

import acm.program.*; 

public class DigitRoot extends ConsoleProgram {

    public void run() {
        println("this program attemts to find the digit root a user enters.");

        int n = readInt("please enter any positive integer: ");

        int dsum = 0;
        int sumtotal = 0;
        int threesum = 0;
        int foursum = 0;

        while (n > 0) {
            dsum += n % 10;
            n /= 10;

            if (dsum > 9) {
                sumtotal = (dsum / 10) + (dsum % 10);
            } else if (sumtotal > 9) {
                threesum = (sumtotal / 10) + (sumtotal % 10);
            } else if (threesum > 9) {
                foursum = (threesum / 10) + (threesum % 10);
            } else if (foursum < 9) {
                println("your digit root is" + foursum);
            } else {
                println("this program is borken.");
            }
        }
    }
}

Answer 1

Example using a recursion – I calculate the sum of the figures by using a String: it is not very efficient but is (slightly) easier. Feel free to improve by using n % 10 in a loop.

public static void main(String[] args) throws Exception {
    System.out.println(calculateRoot(5635)); //prints 1
}

public static int calculateRoot(int n) {
    int sum = sumFigures(n);
    return sum < 10 ? sum : calculateRoot(sum);
}

public static int sumFigures(int n) {
    String s = String.valueOf(n);
    int sum = 0;
    for (char c : s.toCharArray()) {
        sum += c - '0';
    }
    return sum;
}