Calculating Digital Root, is there a better way ?

Question

Hello

This is how i calculated the digital root of an integer.

---

import acm.program.*;

public class Problem7 extends ConsoleProgram
{
    public void run()
    {
        println("This program calculates the digital root of an interger.");

        int num = readInt("Enter the number: ");
        int sum = 0;
        while (true)
        {
            if (num > 0)
            {
                int dsum = num % 10;
                num /= 10;
                sum += dsum;
            }
            else if (sum > 9)
            {
                int dsum = sum % 10;
                sum /= 10;
                sum += dsum;

            } else if (sum <= 9 ) break;
        }
        println("Digital Root is: " + sum);
    }

---

The program works fine.

Is there a better/shorter way of calculating the digital root of a number. ?

---

EDIT/ADDED : Here is the implementation of the above problem by using Tyler's answer, it works as well:

---

import acm.program.*;

public class Problem7 extends ConsoleProgram
{
    public void run()
    {
        println("This program calculates the digital root of an interger.");

        int num = readInt("Enter the number: ");
        println("Digital Root of " + num + " is: " + (1 + (num - 1) % 9));
    }
}

---

Answer 1

I would take the input in as a String instead. This way, you can simply loop through the String and use Integer.parseInt() to grab each number and add them. You can convert that number again to a String and loop through the code to obtain your digital root.

public void run()
{
    println("This program calculates the digital root of an interger.");

    String num = readLine("Enter the number: ");
    int sum = 10;
    while (num > 9) {
      for (int x = 0; x < num.length(); x++) {
         sum = Integer.parseInt(num.charAt(x));
      }
      num = Integer.toString(sum);
   }
   println("Digital Root is: " + sum);
}

Answer 2

Personally, I do not like your loop, which is essentially two loops (first going over the original digits, then going over the digits of sum) mashed into one. How about a sprinkling of recursion:

private int sumDigits(int in){
   if (i>10)
      return in%10 + sumDigits(in/10);
    return in;
}

private int digitalRoot(int in){
    assert (in > 0) ;
    while (in > 9)  in=sumDigits(in);
    return in;
}

Answer 3

#include <stdio.h>

int main(void)
{
   int number;
   scanf("%d", &number);

   printf("The digital root of %d is %d.", number, (1 + (number - 1) % 9));
}

Had I not been able to find Ramans' formula this is how I would write this program...:

#include <stdio.h>
#include <ctype.h>

int main(void)
{
    int c;
    int number = 0;
    while ((c = getchar()) != EOF)
    {
     if (isdigit(c))
      number += (c - '0');
    }
    if (number <= 9)
    {
     printf("The digital root is %d\n", number);
    }
    else
    {
     printf("%d", number);
    }

}

After compiling, to run this, basically you just chain these together. I believe four is the most you could possibly need for an integer.

$ echo 829382938 | ./digitalroot | ./digitalroot | ./digitalroot | ./digitalroot