Maximum possible prime divisors that can exist in numbers having exactly N divisors

Given an integer N which denotes the number of divisors of any number, the task is to find the maximum prime divisors that are possible in number having N divisors.

Examples:

Input: N = 4 
Output: 2

Input: N = 8 
Output: 3 

Naive Approach: In this approach, the idea is to generate all the numbers having exactly N divisors and check for the maximum number of prime divisors. Below are the steps:

1. Define a function is_prime(num) that takes an integer as input and returns True if it is prime, and False otherwise.
2. Check if the number is less than 2, in which case it is not prime.
3. Use a loop to check if the number is divisible by any integer from 2 up to its square root.
4. If it is divisible by any integer, return False.
5. If the loop completes without finding a divisor, return True.

Implementation:

C++

#include <iostream>
#include <cmath>

// Function to check if a number num is prime or not
bool isPrime(int num) {
    if (num < 2) {
        return false;
    }
    for (int i = 2; i * i <= num; i++) {
        if (num % i == 0) {
            return false;
        }
    }
    return true;
}

// Function to count the number of prime divisors in num
int countPrimes(int num) {
    int count = 0;
    for (int i = 2; i * i <= num; i++) {
        if (num % i == 0) {
            if (isPrime(i)) {
                count++;
            }
            if (isPrime(num / i)) {
                count++;
            }
        }
    }
    return count;
}

// Function to find the maximum prime divisor of the number n
int maxPrimeDivisorsBruteForce(int n) {
    int maxPrimes = 0;
    int maxNum = 0;
    for (int num = 2; num < pow(10, n); num++) {
        int divisors = 0;
        for (int i = 1; i <= num; i++) {
            if (num % i == 0) {
                divisors++;
            }
        }
        if (divisors == n) {
            int primeCount = countPrimes(num);
            if (primeCount > maxPrimes) {
                maxPrimes = primeCount;
                maxNum = num;
            }
        }
    }
    return maxPrimes;
}

int main() {
    int n = 4;
    int result = maxPrimeDivisorsBruteForce(n);
    std::cout << result << std::endl;
    return 0;
}

Java

public class MaxPrimeDivisorsBruteForce {

    // Function to check if a number num is prime or not
    static boolean isPrime(int num) {
        if (num < 2) {
            return false;
        }
        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }

    // Function to count the number of prime divisors in num
    static int countPrimes(int num) {
        int count = 0;
        for (int i = 2; i * i <= num; i++) {
            if (num % i == 0) {
                if (isPrime(i)) {
                    count++;
                }
                if (isPrime(num / i)) {
                    count++;
                }
            }
        }
        return count;
    }

    // Function to find the maximum prime divisor of the number n
    static int maxPrimeDivisorsBruteForce(int n) {
        int maxPrimes = 0;
        int maxNum = 0;
        for (int num = 2; num < Math.pow(10, n); num++) {
            int divisors = 0;
            for (int i = 1; i <= num; i++) {
                if (num % i == 0) {
                    divisors++;
                }
            }
            if (divisors == n) {
                int primeCount = countPrimes(num);
                if (primeCount > maxPrimes) {
                    maxPrimes = primeCount;
                    maxNum = num;
                }
            }
        }
        return maxPrimes;
    }

    public static void main(String[] args) {
        int n = 4;
        int result = maxPrimeDivisorsBruteForce(n);
        System.out.println(result);
    }
}

Python3

def is_prime(num):
    """Function to check if a number num is prime or not"""
    if num < 2:
        return False
    for i in range(2, int(num**0.5) + 1):
        if num % i == 0:
            return False
    return True

def count_primes(num):
    """Function to count the number of prime divisors in num"""
    count = 0
    for i in range(2, int(num**0.5) + 1):
        if num % i == 0:
            while num % i == 0:
                num //= i  # Reduce num to its prime factor
            count += 1
    if num > 1:
        count += 1  # If num is a prime number
    return count

def max_prime_divisors_optimized(n):
    """Function to find the maximum prime divisor of the number n"""
    max_primes = 0
    max_num = 0
    for num in range(2, 10**n):
        divisors = 0
        if num % 2 == 0:
            continue  # Skip even numbers as they have more divisors

        for i in range(1, int(num**0.5) + 1):
            if num % i == 0:
                divisors += 1
                if i != num // i:  # Avoid counting twice for perfect squares
                    divisors += 1

        if divisors == n:
            prime_count = count_primes(num)
            if prime_count > max_primes:
                max_primes = prime_count
                max_num = num

    return max_primes

def main():
    n = 4
    result = max_prime_divisors_optimized(n)
    print(result)

if __name__ == "__main__":
    main()

C#

using System;

public class MaxPrimeDivisorsBruteForce
{
    // Function to check if a number num is prime or not
    static bool IsPrime(int num)
    {
        if (num < 2)
        {
            return false;
        }
        for (int i = 2; i * i <= num; i++)
        {
            if (num % i == 0)
            {
                return false;
            }
        }
        return true;
    }

    // Function to count the number of prime divisors in num
    static int CountPrimes(int num)
    {
        int count = 0;
        for (int i = 2; i * i <= num; i++)
        {
            if (num % i == 0)
            {
                if (IsPrime(i))
                {
                    count++;
                }
                if (IsPrime(num / i))
                {
                    count++;
                }
            }
        }
        return count;
    }

    // Function to find the maximum prime divisor of the number n
    static int FindMaxPrimeDivisor(int n)
    {
        int maxPrimes = 0;
        for (int num = 2; num < Math.Pow(10, n); num++)
        {
            int divisors = 0;
            for (int i = 1; i <= num; i++)
            {
                if (num % i == 0)
                {
                    divisors++;
                }
            }
            if (divisors == n)
            {
                int primeCount = CountPrimes(num);
                if (primeCount > maxPrimes)
                {
                    maxPrimes = primeCount;
                }
            }
        }
        return maxPrimes;
    }

    public static void Main(string[] args)
    {
        int n = 4;
        int result = FindMaxPrimeDivisor(n);
        Console.WriteLine(result);
    }
}

Javascript

// Function to check if a number num is prime or not
function isPrime(num) {
    if (num < 2) {
        return false;
    }
    for (let i = 2; i * i <= num; i++) {
        if (num % i === 0) {
            return false;
        }
    }
    return true;
}

// Function to count the number of prime divisors in num
function countPrimes(num) {
    let count = 0;
    for (let i = 2; i * i <= num; i++) {
        if (num % i === 0) {
            if (isPrime(i)) {
                count++;
            }
            if (isPrime(num / i)) {
                count++;
            }
        }
    }
    return count;
}

// Function to find the maximum prime divisor of the number n
function maxPrimeDivisorsBruteForce(n) {
    let maxPrimes = 0;
    let maxNum = 0;
    for (let num = 2; num < Math.pow(10, n); num++) {
        let divisors = 0;
        for (let i = 1; i <= num; i++) {
            if (num % i === 0) {
                divisors++;
            }
        }
        if (divisors === n) {
            let primeCount = countPrimes(num);
            if (primeCount > maxPrimes) {
                maxPrimes = primeCount;
                maxNum = num;
            }
        }
    }
    return maxPrimes;
}

const n = 4;
const result = maxPrimeDivisorsBruteForce(n);
console.log(result);

Output

2

Time Complexity: O(N² * log(N))
Space Complexity: O(N)

Approach: The idea is to find the prime factorization of the number N, then the sum of the powers of the prime divisors is the maximum possible prime divisors of a number can have with N divisors.

For Example:

Let the number of divisors of number be 4,
Then the possible numbers can be 6, 10, 15,...
Divisors of 6 = 1, 2, 3, 6
Total number of prime-divisors = 2 (2, 3)
Prime Factorization of 4 = 22
Sum of powers of prime factors = 2

Below is the implementation of the above approach: 

C++

// C++ implementation to find the
// maximum possible prime divisor
// of a number can have N divisors

#include <iostream>

using namespace std;

#define ll long long int

// Function to find the
// maximum possible prime divisors
// of a number can have with N divisors
void findMaxPrimeDivisor(int n){

    int max_possible_prime = 0;

    // Number of time number
    // divided by 2
    while (n % 2 == 0) {
        max_possible_prime++;
        n = n / 2;
    }

    // Divide by other prime numbers
    for (int i = 3; i * i <= n; i = i + 2) {
        while (n % i == 0) {
            max_possible_prime++;
            n = n / i;
        }
    }

    // If the last number of also
    // prime then also include it
    if (n > 2) {
        max_possible_prime++;
    }

    cout << max_possible_prime << "\n";
}

// Driver Code
int main()
{

    int n = 4;

    // Function Call
    findMaxPrimeDivisor(n);
    return 0;
}

Java

// Java implementation to find the
// maximum possible prime divisor
// of a number can have N divisors
import java.util.*;

class GFG{

// Function to find the
// maximum possible prime divisors
// of a number can have with N divisors
static void findMaxPrimeDivisor(int n)
{
    int max_possible_prime = 0;

    // Number of time number
    // divided by 2
    while (n % 2 == 0)
    {
        max_possible_prime++;
        n = n / 2;
    }

    // Divide by other prime numbers
    for(int i = 3; i * i <= n; i = i + 2)
    {
       while (n % i == 0)
       {
           max_possible_prime++;
           n = n / i;
       }
    }

    // If the last number of also
    // prime then also include it
    if (n > 2)
    {
        max_possible_prime++;
    }
    System.out.print(max_possible_prime + "\n");
}

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

    // Function Call
    findMaxPrimeDivisor(n);
}
}

// This code is contributed by amal kumar choubey

Python3

# Python3 implementation to find the
# maximum possible prime divisor
# of a number can have N divisors

# Function to find the maximum
# possible prime divisors of a
# number can have with N divisors
def findMaxPrimeDivisor(n):

    max_possible_prime = 0

    # Number of time number
    # divided by 2
    while (n % 2 == 0):
        max_possible_prime += 1
        n = n // 2

    # Divide by other prime numbers
    i = 3
    while(i * i <= n):
        while (n % i == 0):

            max_possible_prime += 1
            n = n // i
        i = i + 2

    # If the last number of also
    # prime then also include it
    if (n > 2):
        max_possible_prime += 1

    print(max_possible_prime)

# Driver Code
n = 4

# Function Call
findMaxPrimeDivisor(n)

# This code is contributed by SHUBHAMSINGH10

C#

// C# implementation to find the
// maximum possible prime divisor
// of a number can have N divisors
using System;

class GFG{

// Function to find the
// maximum possible prime divisors
// of a number can have with N divisors
static void findMaxPrimeDivisor(int n)
{
    int max_possible_prime = 0;

    // Number of time number
    // divided by 2
    while (n % 2 == 0)
    {
        max_possible_prime++;
        n = n / 2;
    }

    // Divide by other prime numbers
    for(int i = 3; i * i <= n; i = i + 2)
    {
       while (n % i == 0)
       {
           max_possible_prime++;
           n = n / i;
       }
    }

    // If the last number of also
    // prime then also include it
    if (n > 2)
    {
        max_possible_prime++;
    }
    Console.Write(max_possible_prime + "\n");
}

// Driver Code
public static void Main(String[] args)
{
    int n = 4;

    // Function Call
    findMaxPrimeDivisor(n);
}
}

// This code is contributed by amal kumar choubey

Javascript

<script>

// JavaScript implementation to find the
// maximum possible prime divisor
// of a number can have N divisors

// Function to find the maximum
// possible prime divisors of a
// number can have with N divisors
function findMaxPrimeDivisor(n)
{
    let max_possible_prime = 0;

    // Number of time number
    // divided by 2
    while (n % 2 == 0)
    {
        max_possible_prime++;
        n = Math.floor(n / 2);
    }

    // Divide by other prime numbers
    for(let i = 3; i * i <= n; i = i + 2)
    {
        while (n % i == 0)
        {
            max_possible_prime++;
            n = Math.floor(n / i);
        }
    }

    // If the last number of also
    // prime then also include it
    if (n > 2)
    {
        max_possible_prime++;
    }
    document.write(max_possible_prime + "\n");
}

// Driver Code
let n = 4;

// Function Call
findMaxPrimeDivisor(n);

// This code is contributed by Surbhi Tyagi.

</script>

Output

2

Time Complexity: O(sqrt(N) * logN )
Auxiliary Space: O(1)