Is it a joke

Python (among many other languages) has native support for large integers, which is handy for working with 6**2014. A simple function can be used to accumulate the right-most digit and chop it off, rinse and repeat.

def go(n):
    ans = 0
    while (n > 0):
        ans = ans + n % 10
        n = n / 10
    return ans

go(6**2014)

Using Common Lisp

>(defun sum-digits (number base) (loop for n = number then q for (q r) = (multiple-value-list (truncate n base)) sum r until (zerop q)))

>(sum-digits (expt 6 2014) 10)

reference site : http://rosettacode.org/wiki/Sum digits of an integer

Written in C#:

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using System;
using System.Numerics;

public class Program
{
    public static void Main()
    {
        System.Numerics.BigInteger i = BigInteger.Pow(6, 2014);
        Console.WriteLine(DigitSum(i));
    }

    static int DigitSum(BigInteger n)
    {
        int sum = 0;

        while (n != 0)
        {
            sum += (int)(n % 10);
            n /= 10;
        }

        return sum;
    }
}