Count Divisors of Factorial

Last Updated : 22 Jul, 2026

Given an integer n, determine the total number of positive divisors of n! (the factorial of n). Return 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 120.

Try It Yourself
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Table of Content

[Naive Approach] Brute Force Approach - O(n!) Time and O(1) Space

The idea is to first compute the factorial of n and then count how many positive integers divide it exactly.

C++ 
#include <iostream>
using namespace std;

// Function to return the total number of divisors of n!
int totalDivisors(int n)
{
    // Compute n!
    long long fact = 1;
    for (int i = 2; i <= n; i++)
        fact *= i;

    // Count the divisors of n!
    int count = 0;
    for (long long i = 1; i <= fact; i++)
    {
        if (fact % i == 0)
            count++;
    }

    return count;
}

int main()
{
    int n = 4;
    cout << totalDivisors(n) << endl;

    return 0;
}
Java 
import java.util.*;

class GFG {

    // Function to return the total number of divisors of n!
    static int totalDivisors(int n)
    {
        // Compute n!
        long fact = 1;
        for (int i = 2; i <= n; i++)
            fact *= i;

        // Count the divisors of n!
        int count = 0;
        for (long i = 1; i <= fact; i++) {
            if (fact % i == 0)
                count++;
        }

        return count;
    }

    public static void main(String[] args)
    {
        int n = 4;
        System.out.println(totalDivisors(n));
    }
}
Python 
def totalDivisors(n):

    # Compute n!
    fact = 1
    for i in range(2, n + 1):
        fact *= i

    # Count the divisors of n!
    count = 0
    for i in range(1, fact + 1):
        if fact % i == 0:
            count += 1

    return count


# Driver Code
if __name__ == "__main__":
    n = 4

    print(totalDivisors(n))
C# 
using System;

class GFG {
    
    // Function to return the total number of divisors of n!
    static int totalDivisors(int n)
    {
        // Compute n!
        long fact = 1;
        for (int i = 2; i <= n; i++)
            fact *= i;

        // Count the divisors of n!
        int count = 0;
        for (long i = 1; i <= fact; i++) {
            if (fact % i == 0)
                count++;
        }

        return count;
    }

    static void Main()
    {
        int n = 4;

        Console.WriteLine(totalDivisors(n));
    }
}
JavaScript 
// Function to return the total number of divisors of n!
function totalDivisors(n)
{
    // Compute n!
    let fact = 1n;
    for (let i = 2n; i <= BigInt(n); i++)
        fact *= i;

    // Count the divisors of n!
    let count = 0;
    for (let i = 1n; i <= fact; i++) {
        if (fact % i === 0n)
            count++;
    }

    return count;
}

// Driver Code

const n = 4;

console.log(totalDivisors(n));

Output
8

[Better Approach] Using Prime Factorization - O(sqrt(n!)) Time and O(log(n!)) Space

Instead of checking every number from 1 to fact(n) to determine whether it is a divisor, we first find the prime factorization of fact(n). Once the exponent of each prime factor is known, the total number of divisors can be computed using the formula (e1+1)(e2+1)⋯⋯, where ei is the exponent of a prime factor.

  • Compute the factorial of n.
  • Find the prime factorization of fact(n) using trial division.
  • Store the exponent of each prime factor in a map.
  • Initialize the answer as 1.
  • Multiply the answer by (exponent + 1) for every prime factor.
  • Return the final answer.
C++ 
#include <iostream>
#include <unordered_map>
using namespace std;

// Function to return the total number of divisors of n!
int totalDivisors(int n)
{
    // Compute n!
    long long fact = 1;
    for (int i = 2; i <= n; i++)
        fact *= i;

    unordered_map<long long, int> primeFreq;

    // Find the prime factorization of n!
    for (long long i = 2; i * i <= fact; i++)
    {

        while (fact % i == 0)
        {
            primeFreq[i]++;
            fact /= i;
        }
    }

    // If a prime factor greater than sqrt(n!) remains
    if (fact > 1)
        primeFreq[fact]++;

    int ans = 1;

    // Calculate the total number of divisors
    for (auto &it : primeFreq)
        ans *= (it.second + 1);

    return ans;
}

int main()
{
    int n = 4;
    cout << totalDivisors(n) << endl;

    return 0;
}
Java 
import java.util.*;

class GFG {

    // Function to return the total number of divisors of n!
    static int totalDivisors(int n)
    {
        // Compute n!
        long fact = 1;
        for (int i = 2; i <= n; i++)
            fact *= i;

        HashMap<Long, Integer> primeFreq = new HashMap<>();

        // Find the prime factorization of n!
        for (long i = 2; i * i <= fact; i++) {

            while (fact % i == 0) {
                primeFreq.put(
                    i, primeFreq.getOrDefault(i, 0) + 1);
                fact /= i;
            }
        }

        // If a prime factor greater than sqrt(n!) remains
        if (fact > 1)
            primeFreq.put(
                fact, primeFreq.getOrDefault(fact, 0) + 1);

        int ans = 1;

        // Calculate the total number of divisors
        for (int exponent : primeFreq.values())
            ans *= (exponent + 1);

        return ans;
    }

    public static void main(String[] args)
    {
        int n = 4;
        System.out.println(totalDivisors(n));
    }
}
Python 
def totalDivisors(n):

    # Compute n!
    fact = 1
    for i in range(2, n + 1):
        fact *= i

    primeFreq = {}

    # Find the prime factorization of n!
    i = 2
    while i * i <= fact:

        while fact % i == 0:
            primeFreq[i] = primeFreq.get(i, 0) + 1
            fact //= i

        i += 1

    # If a prime factor greater than sqrt(n!) remains
    if fact > 1:
        primeFreq[fact] = primeFreq.get(fact, 0) + 1

    ans = 1

    # Calculate the total number of divisors
    for exponent in primeFreq.values():
        ans *= (exponent + 1)

    return ans


# Driver Code
if __name__ == "__main__":
    n = 4

    print(totalDivisors(n))
C# 
using System;
using System.Collections.Generic;

class GFG {
    
    // Function to return the total number of divisors of n!
    static int totalDivisors(int n)
    {
        // Compute n!
        long fact = 1;
        for (int i = 2; i <= n; i++)
            fact *= i;

        Dictionary<long, int> primeFreq
            = new Dictionary<long, int>();

        // Find the prime factorization of n!
        for (long i = 2; i * i <= fact; i++) {
            while (fact % i == 0) {
                if (!primeFreq.ContainsKey(i))
                    primeFreq[i] = 0;

                primeFreq[i]++;
                fact /= i;
            }
        }

        // If a prime factor greater than sqrt(n!) remains
        if (fact > 1) {
            if (!primeFreq.ContainsKey(fact))
                primeFreq[fact] = 0;

            primeFreq[fact]++;
        }

        int ans = 1;

        // Calculate the total number of divisors
        foreach(var item in primeFreq) ans
            *= (item.Value + 1);

        return ans;
    }

    static void Main()
    {
        int n = 4;

        Console.WriteLine(totalDivisors(n));
    }
}
JavaScript 
// Function to return the total number of divisors of n!
function totalDivisors(n)
{
    // Compute n!
    let fact = 1n;

    for (let i = 2n; i <= BigInt(n); i++)
        fact *= i;

    let primeFreq = new Map();

    // Find the prime factorization of n!
    for (let i = 2n; i * i <= fact; i++) {

        while (fact % i === 0n) {
            primeFreq.set(i, (primeFreq.get(i) || 0) + 1);
            fact /= i;
        }
    }

    // If a prime factor greater than sqrt(n!) remains
    if (fact > 1n)
        primeFreq.set(fact, (primeFreq.get(fact) || 0) + 1);

    let ans = 1;

    // Calculate the total number of divisors
    for (let exponent of primeFreq.values())
        ans *= (exponent + 1);

    return ans;
}

// Driver Code

const n = 4;

console.log(totalDivisors(n));

Output
8

Note: This approach works correctly only for small values of n. In languages such as C++, Java, and C#, 21! exceeds the range of 64-bit integers, leading to overflow. Although Python (int) and JavaScript (BigInt) can handle larger factorials, explicitly computing and factorizing n! is still inefficient. Therefore, the expected approach using Legendre's Formula is recommended.

[Expected Approach] Using Legendre’s Formula and Sieve of Eratosthenes - O(n log(log n)) Time and O(n) Space

Instead of computing fact(n) explicitly, we directly compute the exponent of every prime in its prime factorization. Using Legendre's Formula, we can efficiently find how many times a prime p appears in fact(n). Once the exponents of all prime factors are known, the total number of divisors can be computed using the formula:

(a1+1)(a2+1)(a3+1)⋯ where a1, a2, a3, ... are the exponents of the prime factors of fact(n).

  • Generate all prime numbers up to n using the Sieve of Eratosthenes.
  • Initialize the answer as 1.
  • Traverse each prime number p from 2 to n.
  • Compute the exponent of p in fact(n) using Legendre's Formula by repeatedly adding ⌊n / p^k⌋ for increasing powers of p.
  • Multiply the answer by (exponent + 1) for every prime.
  • Return the final answer, which represents the total number of divisors of fact(n).
C++ 
#include <iostream>
#include <vector>
using namespace std;

// Function to return the total number of divisors of n!
int totalDivisors(int n)
{
    // Mark all numbers as prime initially.
    vector<bool> isPrime(n + 1, true);

    isPrime[0] = isPrime[1] = false;

    // Generate all prime numbers up to n using
    // the Sieve of Eratosthenes.
    for (int i = 2; i * i <= n; i++)
    {
        if (isPrime[i])
        {
            for (int j = i * i; j <= n; j += i)
                isPrime[j] = false;
        }
    }

    int ans = 1;

    // For every prime p, compute its exponent in n!
    // using Legendre's Formula.
    for (int p = 2; p <= n; p++)
    {

        if (!isPrime[p])
            continue;

        int exponent = 0;
        long long power = p;

        // Count the exponent of p in n!
        while (power <= n)
        {
            exponent += n / power;

            // Prevent overflow while computing p^k.
            if (power > n / p)
                break;

            power *= p;
        }

        // Multiply by (exponent + 1) according to
        // the divisor count formula.
        ans *= (exponent + 1);
    }

    return ans;
}

int main()
{
    int n = 4;
    cout << totalDivisors(n) << endl;

    return 0;
}
Java 
import java.util.*;

class GFG {

    // Function to return the total number of divisors of n!
    static int totalDivisors(int n)
    {
        // Mark all numbers as prime initially.
        boolean[] isPrime = new boolean[n + 1];
        Arrays.fill(isPrime, true);

        isPrime[0] = false;
        isPrime[1] = false;

        // Generate all prime numbers up to n using
        // the Sieve of Eratosthenes.
        for (int i = 2; i * i <= n; i++) {
            if (isPrime[i]) {
                for (int j = i * i; j <= n; j += i)
                    isPrime[j] = false;
            }
        }

        int ans = 1;

        // For every prime p, compute its exponent in n!
        // using Legendre's Formula.
        for (int p = 2; p <= n; p++) {

            if (!isPrime[p])
                continue;

            int exponent = 0;
            long power = p;

            // Count the exponent of p in n!
            while (power <= n) {
                exponent += n / power;

                // Prevent overflow while computing p^k.
                if (power > n / p)
                    break;

                power *= p;
            }

            // Multiply by (exponent + 1) according to
            // the divisor count formula.
            ans *= (exponent + 1);
        }

        return ans;
    }

    public static void main(String[] args)
    {
        int n = 4;

        System.out.println(totalDivisors(n));
    }
}
Python 
# Function to return the total number of divisors of n!
def totalDivisors(n):

    # Mark all numbers as prime initially.
    isPrime = [True] * (n + 1)

    isPrime[0] = False
    isPrime[1] = False

    # Generate all prime numbers up to n using
    # the Sieve of Eratosthenes.
    i = 2
    while i * i <= n:
        if isPrime[i]:
            j = i * i
            while j <= n:
                isPrime[j] = False
                j += i
        i += 1

    ans = 1

    # For every prime p, compute its exponent in n!
    # using Legendre's Formula.
    for p in range(2, n + 1):

        if not isPrime[p]:
            continue

        exponent = 0
        power = p

        # Count the exponent of p in n!
        while power <= n:
            exponent += n // power
            power *= p

        # Multiply by (exponent + 1) according to
        # the divisor count formula.
        ans *= (exponent + 1)

    return ans


# Driver Code
if __name__ == "__main__":
    n = 4

    print(totalDivisors(n))
C# 
using System;

class GFG {
    
    // Function to return the total number of divisors of n!
    static int totalDivisors(int n)
    {
        // Mark all numbers as prime initially.
        bool[] isPrime = new bool[n + 1];

        for (int i = 0; i <= n; i++)
            isPrime[i] = true;

        isPrime[0] = false;
        isPrime[1] = false;

        // Generate all prime numbers up to n using
        // the Sieve of Eratosthenes.
        for (int i = 2; i * i <= n; i++) {
            if (isPrime[i]) {
                for (int j = i * i; j <= n; j += i)
                    isPrime[j] = false;
            }
        }

        int ans = 1;

        // For every prime p, compute its exponent in n!
        // using Legendre's Formula.
        for (int p = 2; p <= n; p++) {
            if (!isPrime[p])
                continue;

            int exponent = 0;
            long power = p;

            // Count the exponent of p in n!
            while (power <= n) {
                exponent += (int)(n / power);

                // Prevent overflow while computing p^k.
                if (power > n / p)
                    break;

                power *= p;
            }

            // Multiply by (exponent + 1) according to
            // the divisor count formula.
            ans *= (exponent + 1);
        }

        return ans;
    }

    static void Main()
    {
        int n = 4;

        Console.WriteLine(totalDivisors(n));
    }
}
JavaScript 
// Function to return the total number of divisors of n!
function totalDivisors(n)
{
    // Mark all numbers as prime initially.
    let isPrime = new Array(n + 1).fill(true);

    isPrime[0] = false;
    isPrime[1] = false;

    // Generate all prime numbers up to n using
    // the Sieve of Eratosthenes.
    for (let i = 2; i * i <= n; i++) {
        if (isPrime[i]) {
            for (let j = i * i; j <= n; j += i)
                isPrime[j] = false;
        }
    }

    let ans = 1;

    // For every prime p, compute its exponent in n!
    // using Legendre's Formula.
    for (let p = 2; p <= n; p++) {

        if (!isPrime[p])
            continue;

        let exponent = 0;
        let power = p;

        // Count the exponent of p in n!
        while (power <= n) {
            exponent += Math.floor(n / power);

            // Prevent overflow while computing p^k.
            if (power > Math.floor(n / p))
                break;

            power *= p;
        }

        // Multiply by (exponent + 1) according to
        // the divisor count formula.
        ans *= (exponent + 1);
    }

    return ans;
}

// Driver Code
let n = 4;

console.log(totalDivisors(n));

Output
8
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