# 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.