Algorithm for digit summing?

Question

I'm searching for an algorithm for Digit summing. Let me outline the basic principle:

Say you have a number: 18268.

1 + 8 + 2 + 6 + 8 = 25

2 + 5 = 7

And 7 is our final number. It's basically adding each number of the whole number until we get down to a single (also known as a 'core') digit. It's often used by numerologists.

I'm searching for an algorithm (doesn't have to be language in-specific) for this. I have searched Google for the last hour with terms such as digit sum algorithm and whatnot but got no suitable results.

Any help would be great, thanks.

Answer 1

I would try this:

int number = 18268;
int core = number;
int total = 0;

while(core > 10)
{
   total = 0;
   number = core;
   while(number > 0)
   {
      total += number % 10;
      number /= 10;
   }

   core = total;
}

Answer 2

1. Mod the whole number by 10.
2. Add the number to an array.
3. Add the whole array.

Answer 3

Googling for 'numerology digit sum algorithm' gives this:

http://wiki.answers.com/Q/Algorithm_to_find_the_sum_of_digits_of_a_given_number

Put a loop around it to sum the result if greater than a single digit, and you're done.

Answer 4

Doesn't work with negative numbers, but I don't know how you would handle it anyhow. You can also change f(x) to be iterative:

sum( x ) =
    while ( ( x = f( x ) ) >= 10 );
    return x;

f( x ) = 
    if ( x >= 10 ) return f( x / 10 ) + x % 10
    return x

You can also take advantage of number theory, giving you this f(x):

f( x ) =
    if ( x == 0 ) return 0
    return x % 9

Answer 5

Because 10-1=9, a little number theory will tell you that the final answer is just n mod 9. Here's code:

ans = n%9;
if(ans==0 && n>0) ans=9; 
return ans;

Example: 18268%9 is 7. (Also see: Casting out nines.)

Answer 6

int number = 18268;
int total = 0;

while(number > 0)
{
   total += number % 10;
   total = total%10;
   number /= 10;
}