Count Divisors of Factorial (original) (raw)

Last Updated : 23 Jul, 2025

Given a number **n, count the total number of divisors of **n!.

**Examples:

**Input : n = 4
**Output: 8
**Explanation:
4! is 24. Divisors of 24 are 1, 2, 3, 4, 6,
8, 12 and 24.

**Input : n = 5
**Output : 16
**Explanation:
5! is 120. Divisors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24 30, 40, 60 and 12

Try It Yourself

A Simple Solution is to first compute the factorial of the given number, then count the number of divisors of the factorial. This solution is not efficient and may cause overflow due to factorial computation.
A better solution is based on Legendre’s formula. Below are the step:

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 value of largest power that divides n! is ?n/p? + ?n/(p2)? + ?n/(p3)? + ......
   Let these values be exp1, exp2, exp3,... Using the above formula, we get the below values for n = 6.
   • 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 (exp1 + 1) * (exp2 + 1) * (exp3 + 1) ... for all prime numbers, For n = 6, the values exp1, exp2, and exp3 are 4 2 and 1 respectively (computed in above step 2). So our result is (4 + 1)*(2 + 1) * (1 + 1) = 30

Below is the implementation of the above idea.

C++ `

// C++ program to find count of divisors in n! #include<bits/stdc++.h> using namespace std; typedef unsigned long long int ull;

// 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 ull factorialDivisors(ull n) { sieve(n); // create sieve

// Initialize result
ull result = 1;

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

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

    // Multiply exponents of all primes less
    // than n
    result = result*(exp+1);
}

// return total divisors
return result;

}

// Driver code int main() { cout << factorialDivisors(6); return 0; }

Java

// JAVA program to find count of divisors in n!

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

// 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];
    
    for(int i=0;i<=n;i++)
    prime[i]=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.add(p);
    }
    
    // Function to find all result of factorial number
    static long factorialDivisors(int n)
    {
        sieve(n); // create sieve
    
        // Initialize result
        long result = 1;
    
        // find exponents of all primes which divides n
        // and less than n
        for (int i=0; i < allPrimes.size(); i++)
        {
            // Current divisor
            long p = allPrimes.get(i);
    
            // Find the highest power (stored in exp)'
            // of allPrimes[i] that divides n using
            // Legendre's formula.
            long exp = 0;
            while (p <= n)
            {
                exp = exp + (n/p);
                p = p*allPrimes.get(i);
            }
    
            // Multiply exponents of all primes less
            // than n
            result = result*(exp+1);
        }
    
        // return total divisors
        return result;
    }
    
    // Driver code
    public static void main(String []args)
    {
        System.out.println(factorialDivisors(6));
        
    }

}

//This code is contributed by ihritik

Python3

Python3 program to find count

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 
        i = p * 2;
        while(i <= n): 
            prime[i] = False;
            i += p;
    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]; 

    # Multiply exponents of all 
    # primes less than n 
    result = result * (exp + 1); 

# return total divisors 
return result; 

Driver Code

print(factorialDivisors(6));

This code is contributed by mits

C#

// C# program to find count 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];
    
    for(int i = 0; i <= n; i++)
    prime[i] = 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.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];
            }
    
            // Multiply exponents of all primes less
            // than n
            result = result * (exp + 1);
        }
    
        // return total divisors
        return result;
    }
    
    // Driver code
    public static void Main()
    {
        Console.WriteLine(factorialDivisors(6));
    }

}

//This code is contributed by chandan_jnu

JavaScript

PHP

`

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

**Auxiliary Space: O(n)

This article is reviewed by team GeeksforGeeks.