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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:

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:

Python3

 `# Python program for the above approach` `# Function to check if a number num is``# prime or not``def` `is_prime(num):``    ``if` `num < ``2``:``        ``return` `False``    ``for` `i ``in` `range``(``2``, ``int``(num ``*``*` `0.5``) ``+` `1``):``        ``if` `num ``%` `i ``=``=` `0``:``            ``return` `False``    ``return` `True` `# Function to count the number of prime``# in the number num``def` `count_primes(num):``    ``count ``=` `0``    ``for` `i ``in` `range``(``2``, ``int``(num ``*``*` `0.5``) ``+` `1``):``        ``if` `num ``%` `i ``=``=` `0``:``            ``if` `is_prime(i):``                ``count ``+``=` `1``            ``if` `is_prime(num ``/``/` `i):``                ``count ``+``=` `1``    ``return` `count` `# Function to find the maximum prime``# divisor of the number n``def` `max_prime_divisors_brute_force(n):``    ``max_primes ``=` `0``    ``max_num ``=` `0``    ``for` `num ``in` `range``(``2``, ``pow``(``10``, n)):``        ``if` `len``([i ``for` `i ``in` `range``(``1``, num ``+` `1``) ``if` `num ``%` `i ``=``=` `0``]) ``=``=` `n:``            ``prime_count ``=` `count_primes(num)``            ``if` `prime_count > max_primes:``                ``max_primes ``=` `prime_count``                ``max_num ``=` `num``    ``return` `max_primes` `# Driver Code``n ``=` `4``result ``=` `max_prime_divisors_brute_force(n)` `print``(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 ` `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

 ``

Output:

`2`

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