Digital sum

Digital root of S is congruent to S mod 9. So no.of no.s less than 1 million which are 1 mod 9 are $\frac{1000000-1}{9}$ = 111111 =S which is congruent to 6 mod 9.

Quite easy!!! The first no. which has digital root of 1 is 1, THEN onwards every 9th no. has a digital root of 1 (Example : 1,10,19,28...) SO..... the last no. is 1000000. it makes up 111111 terms in all..
So the digital root of this no. is 6.. _:) _

Here is my solution in Matlab

function main

clear all close all clc

s=0; for i = 1:999999 if(digitalroot(i)==1) s=s+1; end end

s

a=digitalroot(s)

end

function y=digitalroot(i) while(i>=10) d(1)=floor(i/1e6); d(2)=mod(floor(i/1e5),1e1); d(3)=floor(mod(i,1e5)/1e4); d(4)=floor(mod(i,1e4)/1e3); d(5)=floor(mod(i,1e3)/1e2); d(6)=floor(mod(i,1e2)/1e1); d(7)=mod(i,1e1); i=sum(d); end y=i; end

Here is my solution in c#

private void GetDigitalSum(double num, ref double num2)
    {
        //digital sum

        double sum = 0;

        for (int r = 0; r <= num.ToString().Length - 1; r++)
        {
            sum += double.Parse(num.ToString().Substring(r, 1));

        }

        if (sum.Equals(1))
        {
            num2 = -8;
            return;
        }
        else
        {
            if (sum > 9)
            {
                GetDigitalSum(sum, ref num2);
            }
            else
            {
                num2 = sum;
            }
        }


    }

  Here is the calling subroutine.

   double no = 1;
        double no2 = 0;
        double sumr = 0;
        double sumc = 0;

        ArrayList ar = new ArrayList();

        while (no < 1000000)
        {
            Application.DoEvents();
            lblCounter.Text = no.ToString();

            GetDigitalSum(no, ref no2);

            if (no2.Equals(-8))
            {
                ar.Add(no);
                sumr += double.Parse(no.ToString());
            }

            no += 1;
        }


        GetDigitalSum(ar.Count, ref no2);

        MessageBox.Show(no2.ToString());

       Answer : 6

Here is a programming solution in Python.

count = 0

def digitalRoot(n): #This function calculates the digital root
    Sum = 0

    while 1:
        while n > 0:
            Sum = Sum + (n%10)
            n = n/10

        if Sum >= 10:
            n = Sum
            Sum = 0
        else:
            return Sum
            break

#Now loop through all numbers less than a million and find those where digital root = 1
for x in xrange(1, 1000000):          
    if digitalRoot(x) == 1:
        count = count + 1

print digitalRoot(count)

The programme returns 6 as the answer.