Summing up digits of a very long binary number?

Question

I was asked by a friend :

If 2^10 = 1024 , we can take 1024 and break and summarize its digits :

1+0+2+4 = 7.

This is easy.

However When the input is 2^30000 ( the input actually is a long string "1000...") --there is no .net type which can hold this value .

So there must be a trick to sum its digits (digits of the decimal value)....

Edited :

Related trick ( for finding 10^20 - 16)

100 = 10^2 (one and two zeros)

10^20 = (one and 20 zeros)

hence:

10^20 - 16 = 18 nines, an eight and four.

18*9+8+4 = 174

But I haven't succeeded converting this solution to my problem.( I tried quite a lot).

*Im tagging this question as .net because I can use string functions , math functions from .net library.*

Question

Is there any trick here which can allow me to sum many many numbers which is the result of x^n ?

What is the trick here ?

Edited #2 : Added the .net2 tag (where biginteger is unavailable) - I'm wondering how I could do it without biginteger.(i'm looking for the hidden trick)

Answer 1

From http://blog.singhanuvrat.com/problems/sum-of-digits-in-ab:

public class Problem_16 {
    public long sumOfDigits(int base, int exp) {
        int numberOfDigits = (int) Math.ceil(exp * Math.log10(base));
        int[] digits = new int[numberOfDigits];
        digits[0] = base;
        int currentExp = 1;

        while (currentExp < exp) {
            currentExp++;
            int carry = 0;
            for (int i = 0; i < digits.length; i++) {
                int num = base * digits[i] + carry;
                digits[i] = num % 10;
                carry = num / 10;
            }
        }

        long sum = 0;
        for (int digit : digits)
            sum += digit;

        return sum;
    }

    public static void main(String[] args) {
        int base = 2;
        int exp = 3000;
        System.out.println(new Problem_16().sumOfDigits(base, exp));
    }
}

c#

public class Problem_16 {
    public long sumOfDigits(int base1, int exp) {
        int numberOfDigits = (int) Math.Ceiling(exp * Math.Log10(base1));
        int[] digits = new int[numberOfDigits];
        digits[0] = base1;
        int currentExp = 1;

        while (currentExp < exp) {
            currentExp++;
            int carry = 0;
            for (int i = 0; i < digits.Length; i++) {
                int num = base1 * digits[i] + carry;
                digits[i] = num % 10;
                carry = num / 10;
            }
        }

        long sum = 0;
        foreach (int digit in  digits)
            sum += digit;

        return sum;
    }
}


void Main()
{
     int base1 = 2;
        int exp = 3000000;
        Console.WriteLine (new Problem_16().sumOfDigits(base1, exp));

}