Sum of Factorial's digit

Python Solution:

import math

print sum(map(int, list(str(math.factorial(2013)))))

Python Solution

from math import factorial

def digit_sum(n):
    sum_ = 0
    while n:
        sum_ += n % 10
        n /= 10
    return sum_

A = factorial(2013)

print digit_sum(A), digit_sum(A) % 1000

Output: $24021, \fbox{21}$

import sys sys.setrecursionlimit(5000) def fact(n): if n == 0 : return 1 else: return n*fact(n-1)

def sum_digits(n): s = 0 while n: s += n % 10 n /= 10 return s

m = fact(2013)

print sum_digits(m)

In the start i have to increase the recursion limit as if we run the code normally there will be run time error. please correct the indentation as i am unable to write the code in a formatted manner.

Python Implementation:

def factorial(n): 
  a=2 
  fact=1 
  while a<=n: 
    fact=fact*a 
    a=a+1 
  return fact

def sum_digits(n): 
  string=str(n) 
  a=0 
  sum=0 
  while a<len(string): 
    sum=sum+int(string[a]) 
    a=a+1 
  return sum 

x=sum_digits(factorial(2013)) 

print x

Result= $24\boxed{021}$