Find sum of digits in factorial of a number

Given a number n, write code to find the sum of digits in the factorial of the number. Given n <= 5000

Examples:

Input : 10
Output : 27

Input : 100
Output : 648

It is not possible to store a number as large as 100! under some data types so, idea is to store extremely large number in vector.



1) Create a vector to store factorial digits and
   initialize it with 1.
2) One by one multiply numbers from 1 to n to 
   the vector. We use school mathematics for this
   purpose.
3) Sum all the elements in vector and return the sum.

C++

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// C++ program to find sum of digits in factorial
// of a number
#include <bits/stdc++.h>
using namespace std;
  
// Function to multiply x with large number
// stored in vector v. Result is stored in v.
void multiply(vector<int> &v, int x)
{
    int carry = 0, res;
    int size = v.size();
    for (int i = 0 ; i < size ; i++)
    {
        // Calculate res + prev carry
        int res = carry + v[i] * x;
  
        // updation at ith position
        v[i] = res % 10;
        carry = res / 10;
    }
    while (carry != 0)
    {
        v.push_back(carry % 10);
        carry /= 10;
    }
}
  
// Returns sum of digits in n!
int findSumOfDigits(int n)
{
    vector<int> v;     // create a vector of type int
    v.push_back(1);    // adds 1 to the end
  
    // One by one multiply i to current vector
    // and update the vector.
    for (int i=1; i<=n; i++)
        multiply(v, i);
  
    // Find sum of digits in vector v[]
    int sum = 0;
    int size = v.size();
    for (int i = 0 ; i < size ; i++)
        sum += v[i];
    return sum;
}
  
// Driver code
int main()
{
    int n = 1000;
    cout << findSumOfDigits(n);
    return 0;
}

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Java

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// Java program to find sum of digits in factorial 
// of a number 
import java.util.*;
class GFG{
// Function to multiply x with large number 
// stored in vector v. Result is stored in v.
static ArrayList<Integer> v=new ArrayList<Integer>();
static void multiply(int x) 
    int carry = 0
    int size = v.size(); 
    for (int i = 0 ; i < size ; i++) 
    
        // Calculate res + prev carry 
        int res = carry + v.get(i) * x; 
  
        // updation at ith position 
        v.set(i,res % 10); 
        carry = res / 10
    
    while (carry != 0
    
        v.add(carry % 10); 
        carry /= 10
    
  
// Returns sum of digits in n! 
static int findSumOfDigits(int n) 
    v.add(1); // adds 1 to the end 
  
    // One by one multiply i to current vector 
    // and update the vector. 
    for (int i=1; i<=n; i++) 
        multiply(i); 
  
    // Find sum of digits in vector v[] 
    int sum = 0
    int size = v.size(); 
    for (int i = 0 ; i < size ; i++) 
        sum += v.get(i); 
    return sum; 
  
// Driver code 
public static void main(String[] args) 
    int n = 1000
    System.out.println(findSumOfDigits(n)); 
}
// this code is contributed by mits

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Python 3

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# Python 3 program to find sum of digits
# in factorial of a number
  
# Function to multiply x with large number
# stored in vector v. Result is stored in v.
def multiply(v, x):
    carry = 0
    size = len(v)
    for i in range(size):
          
        # Calculate res + prev carry
        res = carry + v[i] * x
  
        # updation at ith position
        v[i] = res % 10
        carry = res // 10
  
    while (carry != 0):
        v.append(carry % 10)
        carry //= 10
  
# Returns sum of digits in n!
def findSumOfDigits( n):
    v = []     # create a vector of type int
    v.append(1) # adds 1 to the end
  
    # One by one multiply i to current 
    # vector and update the vector.
    for i in range(1, n + 1):
        multiply(v, i)
  
    # Find sum of digits in vector v[]
    sum = 0
    size = len(v)
    for i in range(size):
        sum += v[i]
    return sum
  
# Driver code
if __name__ == "__main__":
      
    n = 1000
    print(findSumOfDigits(n))
  
# This code is contributed by ita_c

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C#

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// C# program to find sum of digits in factorial 
// of a number 
using System;
using System.Collections;
class GFG{
// Function to multiply x with large number 
// stored in vector v. Result is stored in v.
static ArrayList v=new ArrayList();
static void multiply(int x) 
    int carry = 0; 
    int size = v.Count; 
    for (int i = 0 ; i < size ; i++) 
    
        // Calculate res + prev carry 
        int res = carry + (int)v[i] * x; 
  
        // updation at ith position 
        v[i] = res % 10; 
        carry = res / 10; 
    
    while (carry != 0) 
    
        v.Add(carry % 10); 
        carry /= 10; 
    
  
// Returns sum of digits in n! 
static int findSumOfDigits(int n) 
    v.Add(1); // adds 1 to the end 
  
    // One by one multiply i to current vector 
    // and update the vector. 
    for (int i=1; i<=n; i++) 
        multiply(i); 
  
    // Find sum of digits in vector v[] 
    int sum = 0; 
    int size = v.Count; 
    for (int i = 0 ; i < size ; i++) 
        sum += (int)v[i]; 
    return sum; 
  
// Driver code 
static void Main() 
    int n = 1000; 
    Console.WriteLine(findSumOfDigits(n)); 
}
// this code is contributed by mits

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PHP

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<?php
// PHP program to find sum of digits in 
// factorial of a number 
  
// Function to multiply x with large number 
// stored in vector v. Result is stored in v. 
function multiply(&$v, $x
    $carry = 0; 
    $size = count($v); 
    for ($i = 0 ; $i < $size ; $i++) 
    
        // Calculate res + prev carry 
        $res = $carry + $v[$i] * $x
  
        // updation at ith position 
        $v[$i] = $res % 10; 
        $carry = (int)($res / 10); 
    
    while ($carry != 0) 
    
        array_push($v, $carry % 10); 
        $carry = (int)($carry / 10); 
    
  
// Returns sum of digits in n! 
function findSumOfDigits($n
    $v = array(); // create a vector of type int 
    array_push($v, 1); // adds 1 to the end 
  
    // One by one multiply i to current vector 
    // and update the vector. 
    for ($i = 1; $i <= $n; $i++) 
        multiply($v, $i); 
  
    // Find sum of digits in vector v[] 
    $sum = 0; 
    $size = count($v); 
    for ($i = 0 ; $i < $size ; $i++) 
        $sum += $v[$i]; 
    return $sum
  
// Driver code 
$n = 1000; 
print(findSumOfDigits($n)); 
  
// This code is contributed by mits 
?>

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Output:

10539

This article is contributed by Sakshi Tiwari . If you like GeeksforGeeks(We know you do!) and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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