Why is the solution like this?

Question

From Java Malik textbook- determine if an number is divisible by 11..

Code Solution provided:

import java.util.*;

public class Divby11
{
    static Scanner console = new Scanner(System.in);

    public static void main (String[] args)
    {
        int num, temp, sum;
        char sign;

        System.out.print("Enter a positive integer: ");
        num = console.nextInt();
        System.out.println();

        temp = num;

        sum = 0;
        sign = '+';

        do
        {
            switch (sign)
            {
            case '+' :
                sum = sum + num % 10;
                sign = '-';
                break;

            case '-' :
                sum = sum - num % 10;
                sign = '+';
            }

            num = num / 10;       //remove the last digit
        }
        while (num > 0);

        if (sum % 11 == 0)
            System.out.println(temp + " is divisible by 11");
        else
            System.out.println(temp + " is not divisible by 11");
    }

Why go through all the effort above and just say...

  if (sum % 11 == 0)
            System.out.println(temp + " is divisible by 11");
        else
            System.out.println(temp + " is not divisible by 11");

Can any of you experts see why the author would do it this way (long way)?

Answer 1

This code example isn't actually dividing by eleven. If you see, it's alternating between adding and subtracting each digit, then checks at the very end if the result is divisible by 11.

For example, look at the following number and how this algorithm works with it:

Start with sum=0, sign='+', num=517
First iteration: sum=7, sign='-', num=51
Second iteration: sum=6, sign='+', num=5
Final iteration: sum=11, sign='-', num=0

The final result is divisible by eleven.

EDIT: The algorithm does indeed look to be implementing the divisibility rule for 11 as dfa mentions in his answer.

Answer 2

for the Divisibility Rule of 11:

• form the alternating sum of the digits
• if this sum is divisible for 11 then the number is divisible for 11

Examples

• 68090 = 0 - 9 + 0 - 8 + 6 = -11 => TRUE
• 493827 = 7 - 2 + 8 - 3 + 9 - 4 = 15 = 4 => FALSE

Answer 3

I suspect it is simulating the manual test that digits in the odd positions and the digits in the even positions differ by a factor of 11. In practice using %11 would be the way to go.

EDIT: If the example were truly trying to avoid doing % 11, it should send the sum through again until it is 0.

Answer 4

You will have to provide more context from the book as to what the author was trying to demonstrate. This code example does not check to see if the number entered is divisible by 11. What it does is it adds every other digit, subtracts every other digit and then checks THAT number to see if it's divisible by 10.

EX Entered number is 4938 It takes the 8 adds it to sum Divides by ten giving 493 Takes the 3 subtracts it from sum: sum = 5 Divides by ten giving 49 Takes 9 and adds it to sum: sum = 14 Divides by ten giving 4 Takes 4 subtracts it from sum: sum = 10

THEN it checks if this is divisible by 11.

Ok, I know why now. He/she's trying to teach you something besides computing about numbers

Answer 5

It an example to show how to implement that particular check. Using your example would not demonstrate the same code methodologies.