Issue

This Content is from Stack Overflow. Question asked by Nutnicha

I have written a code which takes in a number and multiplies it’s digits together and here it is.

public class Happy
    {   public static int findSum(int k)
        {
            int sum = 0;
            while (k != 0)
            {
                sum = sum + ((k % 10)*(k % 10));
                k = k/10;
            }
            return sum;
        }
        public static void main(String args[])
        {
            System.out.println(findSum(713));
        }
    }

and my code returns 19 when my input is 82, because 1^2 + 9^2 is 82.

However, I want to modify my code so that my code would continue to square each digit and add them together until there is only one non-zero digit left.

So basically, if my input is 19, the code would return 1 because:

1^2 + 9^2 = 82, from which digits 8 and 2 would do: 8^2 + 2^2 = 100, from which 1^2 + 0^2 + 0^2 = 1.

What changes to my code should I make in order for it to work?

Solution

You could call your method recursivly by checking if sum > 9

public static int findSum(int k)
{
    int sum = 0;
    while (k != 0)
    {
        sum = sum + ((k % 10)*(k % 10));
        k = k/10;
    }
    return sum > 9 ? findSum(sum) : sum;
}

This Question was asked in StackOverflow by Nutnicha and Answered by Eritrean It is licensed under the terms of CC BY-SA 2.5. - CC BY-SA 3.0. - CC BY-SA 4.0.