Sum of divisors of factorial of a number (original) (raw)

Last Updated : 23 Jul, 2025

Given a number n, we need to calculate the sum of divisors of factorial of the number.

Examples:

Input : 4 Output : 60 Factorial of 4 is 24. Divisors of 24 are 1 2 3 4 6 8 12 24, sum of these is 60.

Input : 6 Output : 2418

A Simple Solution is to first compute the factorial of the given number, then count the number divisors of the factorial. This solution is not efficient and may cause overflow due to factorial computation.

Below is the implementation of the above approach:

C++ `

// C++ program to find sum of proper divisor of // factorial of a number #include <bits/stdc++.h> using namespace std;

// function to calculate factorial int fact(int n) { if (n == 0) return 1; return n * fact(n - 1); }

// function to calculate sum of divisor int div(int x) { int ans = 0; for (int i = 1; i<= x; i++) if (x % i == 0) ans += i; return ans; }

// Returns sum of divisors of n! int sumFactDiv(int n) { return div(fact(n)); }

// Driver Code int main() { int n = 4; cout << sumFactDiv(n); }

// This code is contributed // by Akanksha Rai

C

// C program to find sum of proper divisor of // factorial of a number #include <stdio.h> // function to calculate factorial

int fact(int n) { if (n == 0) return 1; return n * fact(n - 1); }

// function to calculate sum of divisor

int div(int x) { int ans = 0; for (int i = 1; i<= x; i++) if (x % i == 0) ans += i; return ans; }

// Returns sum of divisors of n!

int sumFactDiv(int n) { return div(fact(n)); }

// driver program

int main() { int n = 4; printf("%d",sumFactDiv(n)); }

Java

// Java program to find sum of proper divisor of // factorial of a number import java.io.; import java.util.;

public class Division { // function to calculate factorial static int fact(int n) { if (n == 0) return 1; return n*fact(n-1); }

// function to calculate sum of divisor
static int div(int x)
{
    int ans = 0;
    for (int i = 1; i<= x; i++)
        if (x%i == 0)
            ans += i;
    return ans;
}

// Returns sum of divisors of n!
static int sumFactDiv(int n)
{
    return div(fact(n));
}

// driver program
public static void main(String args[])
{
    int n = 4;
    System.out.println(sumFactDiv(n));
}

}

Python3

Python 3 program to find sum of proper

divisor of factorial of a number

function to calculate factorial

def fact(n):

if (n == 0):
    return 1
return n * fact(n - 1)

function to calculate sum

of divisor

def div(x): ans = 0; for i in range(1, x + 1): if (x % i == 0): ans += i return ans

Returns sum of divisors of n!

def sumFactDiv(n): return div(fact(n))

Driver Code

n = 4 print(sumFactDiv(n))

This code is contributed

by Rajput-Ji

C#

// C# program to find sum of proper // divisor of factorial of a number using System; class Division {

// function to calculate factorial
static int fac(int n)
{
    if (n == 0)
        return 1;
    return n * fac(n - 1);
}

// function to calculate
// sum of divisor
static int div(int x)
{
    int ans = 0;
    for (int i = 1; i <= x; i++)
        if (x % i == 0)
            ans += i;
    return ans;
}

// Returns sum of divisors of n!
static int sumFactDiv(int n)
{
    return div(fac(n));
}

// Driver Code
public static void Main()
{
    int n = 4;
    Console.Write(sumFactDiv(n));
}

}

// This code is contributed by Nitin Mittal.

PHP

n∗fact(n * fact(n∗fact(n - 1); } // function to calculate sum of divisor function div($x) { $ans = 0; for ($i = 1; i<=i<= i<=x; $i++) if ($x % $i == 0) ans+=ans += ans+=i; return $ans; } // Returns sum of divisors of n! function sumFactDiv($n) { return div(fact($n)); } // Driver Code $n = 4; echo sumFactDiv($n); // This code is contributed // by Akanksha Rai ?>

JavaScript

`

Output :

60

Time Complexity: O(n!)

Auxiliary Space: O(1)

An efficient solution is based on Legendre’s formula. Below are the steps.

  1. Find all prime numbers less than or equal to n (input number). We can use Sieve Algorithm for this. Let n be 6. All prime numbers less than 6 are {2, 3, 5}.
  2. For each prime number, p find the largest power of it that divides n!. We use Legendre’s formula for this purpose.
    • The largest power of 2 that divides 6!, exp1 = 4.
    • The largest power of 3 that divides 6!, exp2 = 2.
    • The largest power of 5 that divides 6!, exp3 = 1.
  3. The result is based on the Divisor Function. C++ `

// C++ program to find sum of divisors in n! #include<bits/stdc++.h> #include<math.h> using namespace std;

// allPrimes[] stores all prime numbers less // than or equal to n. vector allPrimes;

// Fills above vector allPrimes[] for a given n void sieve(int n) { // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // not a prime, else true. vector prime(n+1, true);

// Loop to update prime[]
for (int p = 2; p*p <= n; p++)
{
    // If prime[p] is not changed, then it
    // is a prime
    if (prime[p] == true)
    {
        // Update all multiples of p
        for (int i = p*2; i <= n; i += p)
            prime[i] = false;
    }
}

// Store primes in the vector allPrimes
for (int p = 2; p <= n; p++)
    if (prime[p])
        allPrimes.push_back(p);

}

// Function to find all result of factorial number int factorialDivisors(int n) { sieve(n); // create sieve

// Initialize result
int result = 1;

// find exponents of all primes which divides n
// and less than n
for (int i = 0; i < allPrimes.size(); i++)
{
    // Current divisor
    int p = allPrimes[i];

    // Find the highest power (stored in exp)'
    // of allPrimes[i] that divides n using
    // Legendre's formula.
    int exp = 0;
    while (p <= n)
    {
        exp = exp + (n/p);
        p = p*allPrimes[i];
    }

    // Using the divisor function to calculate
    // the sum
    result = result*(pow(allPrimes[i], exp+1)-1)/
                                (allPrimes[i]-1);
}

// return total divisors
return result;

}

// Driver program to run the cases int main() { cout << factorialDivisors(4); return 0; }

Java

// Java program to find sum of divisors in n! import java.util.*;

class GFG{ // allPrimes[] stores all prime numbers less // than or equal to n. static ArrayList allPrimes=new ArrayList();

// Fills above vector allPrimes[] for a given n static void sieve(int n) { // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // not a prime, else true. boolean[] prime=new boolean[n+1]; 

// Loop to update prime[] 
for (int p = 2; p*p <= n; p++) 
{ 
    // If prime[p] is not changed, then it 
    // is a prime 
    if (prime[p] == false) 
    { 
        // Update all multiples of p 
        for (int i = p*2; i <= n; i += p) 
            prime[i] = true; 
    } 
} 

// Store primes in the vector allPrimes 
for (int p = 2; p <= n; p++) 
    if (prime[p]==false) 
        allPrimes.add(p);

}

// Function to find all result of factorial number static int factorialDivisors(int n) { sieve(n); // create sieve 

// Initialize result 
int result = 1; 

// find exponents of all primes which divides n 
// and less than n 
for (int i = 0; i < allPrimes.size(); i++) 
{ 
    // Current divisor 
    int p = allPrimes.get(i); 

    // Find the highest power (stored in exp)' 
    // of allPrimes[i] that divides n using 
    // Legendre's formula. 
    int exp = 0; 
    while (p <= n) 
    { 
        exp = exp + (n/p); 
        p = p*allPrimes.get(i); 
    } 

    // Using the divisor function to calculate 
    // the sum 
    result = result*((int)Math.pow(allPrimes.get(i), exp+1)-1)/ 
                                (allPrimes.get(i)-1); 
} 

// return total divisors 
return result;

}

// Driver program to run the cases public static void main(String[] args) { System.out.println(factorialDivisors(4)); } } // This code is contributed by mits

Python3

Python3 program to find sum of divisors in n!

allPrimes[] stores all prime numbers

less than or equal to n.

allPrimes = [];

Fills above vector allPrimes[]

for a given n

def sieve(n): 

# Create a boolean array "prime[0..n]" 
# and initialize all entries it as true. 
# A value in prime[i] will finally be 
# false if i is not a prime, else true. 
prime = [True] * (n + 1); 

# Loop to update prime[]
p = 2;
while (p * p <= n): 
    
    # If prime[p] is not changed, 
    # then it is a prime 
    if (prime[p] == True): 
        
        # Update all multiples of p 
        for i in range(p * 2, n + 1, p): 
            prime[i] = False; 
    p += 1; 

# Store primes in the vector allPrimes 
for p in range(2, n + 1): 
    if (prime[p]): 
        allPrimes.append(p);

Function to find all result of factorial number

def factorialDivisors(n): 

sieve(n); # create sieve 

# Initialize result 
result = 1; 

# find exponents of all primes which 
# divides n and less than n 
for i in range(len(allPrimes)): 
    
    # Current divisor 
    p = allPrimes[i]; 

    # Find the highest power (stored in exp)' 
    # of allPrimes[i] that divides n using 
    # Legendre's formula. 
    exp = 0; 
    while (p <= n):
        exp = exp + int(n / p); 
        p = p * allPrimes[i]; 

    # Using the divisor function to  
    # calculate the sum 
    result = int(result * (pow(allPrimes[i], exp + 1) - 1) / 
                                       (allPrimes[i] - 1)); 

# return total divisors 
return result;

Driver Code

print(factorialDivisors(4));

This code is contributed by mits

C#

// C# program to find sum of divisors in n! using System; using System.Collections;

class GFG{ // allPrimes[] stores all prime numbers less // than or equal to n. static ArrayList allPrimes=new ArrayList();

// Fills above vector allPrimes[] for a given n static void sieve(int n) { // Create a boolean array "prime[0..n]" and // initialize all entries it as true. A value // in prime[i] will finally be false if i is // not a prime, else true. bool[] prime=new bool[n+1]; 

// Loop to update prime[] 
for (int p = 2; p*p <= n; p++) 
{ 
    // If prime[p] is not changed, then it 
    // is a prime 
    if (prime[p] == false) 
    { 
        // Update all multiples of p 
        for (int i = p*2; i <= n; i += p) 
            prime[i] = true; 
    } 
} 

// Store primes in the vector allPrimes 
for (int p = 2; p <= n; p++) 
    if (prime[p]==false) 
        allPrimes.Add(p);

}

// Function to find all result of factorial number static int factorialDivisors(int n) { sieve(n); // create sieve 

// Initialize result 
int result = 1; 

// find exponents of all primes which divides n 
// and less than n 
for (int i = 0; i < allPrimes.Count; i++) 
{ 
    // Current divisor 
    int p = (int)allPrimes[i]; 

    // Find the highest power (stored in exp)' 
    // of allPrimes[i] that divides n using 
    // Legendre's formula. 
    int exp = 0; 
    while (p <= n) 
    { 
        exp = exp + (n/p); 
        p = p*(int)allPrimes[i]; 
    } 

    // Using the divisor function to calculate 
    // the sum 
    result = result*((int)Math.Pow((int)allPrimes[i], exp+1)-1)/ 
                                ((int)allPrimes[i]-1); 
} 

// return total divisors 
return result;

}

// Driver program to run the cases static void Main() { Console.WriteLine(factorialDivisors(4)); } } // This code is contributed by mits

PHP

prime=arrayfill(0,prime=array_fill(0,prime=arrayf​ill(0,n+1,true); // Loop to update prime[] for ($p = 2; p∗p*p∗p <= n;n; n;p++) { // If prime[p] is not changed, then it // is a prime if ($prime[$p] == true) { // Update all multiples of p for ($i = p∗2;p*2; p∗2;i <= n;n; n;i += $p) prime[prime[prime[i] = false; } } // Store primes in the vector allPrimes for ($p = 2; p<=p <= p<=n; $p++) if ($prime[$p]) array_push($allPrimes,$p); } // Function to find all result of factorial number function factorialDivisors($n) { global $allPrimes; sieve($n); // create sieve // Initialize result $result = 1; // find exponents of all primes which divides n // and less than n for ($i = 0; i<count(i < count(i<count(allPrimes); $i++) { // Current divisor p=p = p=allPrimes[$i]; // Find the highest power (stored in exp)' // of allPrimes[i] that divides n using // Legendre's formula. $exp = 0; while ($p <= $n) { exp=exp = exp=exp + (int)($n/$p); p=p = p=p*$allPrimes[$i]; } // Using the divisor function to calculate // the sum result=result = result=result*(pow($allPrimes[$i], $exp+1)-1)/ ($allPrimes[$i]-1); } // return total divisors return $result; } // Driver program to run the cases print(factorialDivisors(4)); // This code is contributed by mits ?>

JavaScript

`

Output:

60

Time Complexity: O(n*log(log(n)))

Auxiliary Space: O(n)