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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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++

// 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; 
} 

Java

// 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 

Python 3

# 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 

C#

// 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 

PHP

<?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  
?> 

Output:

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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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python 3

C#

PHP

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