如果整数可被3整除，则该方法返回true；如果整数不能被3整除，则该方法返回false。

Question

This is what I have so far; I have to use this main method.

public class HW4 {

    public static boolean isDivisibleByThree(String n) {
        int sum = 0;
        int value;

        for (int k = 0; k < n.length(); k++) {
            char ch = n.charAt(k);
            value = Character.getNumericValue(ch);
            sum = sum*value;
        }
        return sum*3 == 0;
    }
}

It always comes out true and I'm really stuck in this part. So if you can, can you help me out?

Answer 1

A sum is a cumulative addition (not multiplication).

Change this line:

sum = sum * value;

To

sum = sum + value;

Or the more brief version:

sum += value;

Answer 2

Two things:

1. sum = sum * value ? This should probably be sum = sum + value , or short sum += value
2. sum * 3 == 0 should probably be sum % 3 == 0

If you are required to not use the % operator, you could alternatively do:

double check = (double)sum / 3.0;
return check == (int)check;

---

The problem with negative numbers is that the - gets parsed too, you could sove it by dropping it:

if (n[0] == '-') {
    n = n.substring(1);
}

This drops the sign if it is negative and does nothing otherwise.

Answer 3

Much easier solution: use the mod-function:

int number = int.Parse(input);
bool result = (number % 3 == 0);

Answer 4

Unless I'm missing something, you would first use Integer.parseInt(String) to parse the int from the String . Then you can divide that value by 3 using integer division. Finally, test if that number multiplied by 3 is the original value.

int value = Integer.parseInt(n);
int third = value / 3;
return (value == third * 3);