Java Problem

[kasimo420]

I have a question with a program. The assignment is "Write a program to verify the statement Numbers whose sum of digits is divisible by 3 represent numbers divisible by 3. Input a 5 digit integer from the keyboard. Find the sum of the digits, call it sum. Verify that either(a) both n and sum are divisible by 3 or (b) both are indivisible by 3.
Your output is:
Given number =
Sum of digits=
One of the following
a. Both number and sum are divisible by 3
b. Both number and sum are indivisible by 3
c. The famous statement is wrong

This is what I have so far..

import java.util.*;
import java.io.*;
class magicruleof3
{
      public static void main(String arg[])
      {
            int number;
            boolean done=false;
            Scanner input=new Scanner(System.in);
            System.out.println("Please input a 5-digit number");
            number = input.nextInt();
              {
                  String numberStr=Integer.toString(number);
                  int sum=0;
                  for(int i=0;i<numberStr.length();i++)
                  {
                        int remainder=number%10;
                        sum+=remainder;
                        number/=10;
                        System.out.println("Sum of the digits:"+sum);
                  }
                  if(number%3==0&&!done);
                  {
                             
                        if(!done)System.out.println("Both n and sum are indivisible by 3");
                        if(!done)System.out.println("The famous statement is wrong 3");
                        System.out.println("Both n and sum are divisible by 3");done=true;
                        
                  }
              }
      }
                  

}

All is working except the sum of the numbers and the output. Any help is greatly appreciated.

[JosAH]

This recursive: if you want to compute the sum of a bunch of digits this sum either equals the single digits if the number < 10; otherwise the total sum is the sum of all the other digits plus the last digit. Here's the code:

int dsum(int n) { // assume n >= 0
   return (n < 10)?n:(n%10+dsum(n/10));
}

I think you can take it from there.

kind regards,

Jos

[kasimo420]

Tried that, not 100% understanding.

[Nepomuk]

OK, there are two things I guess you may not understand about what Jos posted.

1. return condition ? a : b;

   is basically a short version of

   if (condition) return a;
   else return b;

2. The recursion. A simple example for recursion is the factorial function. What it does mathematically is this:
   The factorial of n (normally written as n!) is n multiplied by all natural numbers smaller than n and bigger than 0 (or 1, that makes no difference). So for example:
   4! = 4 * 3 * 2 * 1 = 24
   You can also write a program to calculate that:
   

   int fac(int n) {
     if(n <= 1) return 1;
     else return n * fac(n-1);
   }

   (We just assume that n >= 0.)
   So, that function will call itself with a (normally) different value. It also needs some value, where it will stop (here anything smaller than or equal to 1).
   Jos' function is a bit more complex, but if you have a good look at it, you should understand what it does.

Greetings,
Nepomuk

[kasimo420]

Dern. Still need help.

[Nepomuk]

With what? You'll have to give us more information.

Greetings,
Nepomuk