Given an integer **X**, denoting the number of factors of a positive integer N can have. the task is to find the maximum number of distinct prime factors the number **N** can have.

**Examples:**

Input:X = 9

Output: 2

Explanation:

Some of the possible numbers having 9 factors are:

256: 1,2, 4, 8, 16, 32, 64, 128, 256

Number of prime factors = 1

36: 1,2, 3, 4, 6, 9, 12, 18, 36

Number of prime factors = 2

Input:X = 8

Output:3

Some of the numbers having 8 factors are:

128 : 1,2,4, 8, 16, 32, 64, 128

Number of prime factors = 1

24 : 1,2, 3,4, 6, 8, 12, 24

Number of prime factors = 2

30 : 1,2, 3, 5, 6, 10, 15, 30

Number of prime factors = 3

**Approach:** The key observation in the problem is, any number can be represented into its prime factors as follows:

// Number can be represented in its // prime factors as follows N = a^p * b^q * c^r .. // Number of prime factors can be // defined as follows Number of Factors = (p+1) * (q+1) * (r+1)..

In the above problem, the number of the prime factors are given which can be used to find the maximum prime factors possible for a number with the given count of factors as follows:

X = (p+1) * (q+1) * (r+1) // So the maximum number of prime // factors of the given number greater // than 1 can lead to a number N. Let's say X = 12 X = 2 * 2 * 3 Then possible N can be: N = a^{(2-1)}* b^{(2-1)}* c^{(3-1)}N = a^{1}* b^{1}* c^{2}// Here a, b, and c can be any distinct prime // numbers to get the possible value of N N = 2^{1}* 3^{1}* 5^{2}N = 150

Therefore, the maximum count of prime divisors of a number can have is the count of the prime factors in the factorization of the count of factors of the number.

Below is the implementation of the above approach:

## C++

`// C++ implmentation to find the ` `// maximum count of the prime factors ` `// by the count of factors of number ` ` ` `#include <iostream> ` `#include <math.h> ` ` ` `using` `namespace` `std; ` ` ` `// Function to count number ` `// of prime factors of x ` `int` `countPrimeFactors(` `int` `n) ` `{ ` ` ` `if` `(n == 1) ` ` ` `return` `0; ` ` ` ` ` `// Count variable is ` ` ` `// incremented form every ` ` ` `// prime factor of x ` ` ` `int` `cnt = 0; ` ` ` `while` `(n % 2 == 0) { ` ` ` `cnt++; ` ` ` `n = n / 2; ` ` ` `} ` ` ` ` ` `// Loop to count the number ` ` ` `// of the prime factors of ` ` ` `// the given number ` ` ` `for` `(` `int` `i = 3; i <= ` `sqrt` `(n); ` ` ` `i += 2) { ` ` ` `while` `(n % i == 0) { ` ` ` `cnt++; ` ` ` `n = n / i; ` ` ` `} ` ` ` `} ` ` ` ` ` `if` `(n > 2) ` ` ` `cnt++; ` ` ` ` ` `return` `cnt; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `x = 8; ` ` ` `int` `prime_factor_cnt = countPrimeFactors(x); ` ` ` `cout << prime_factor_cnt << endl; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implmentation to find the ` `// maximum count of the prime factors ` `// by the count of factors of number ` `class` `GFG{ ` ` ` `// Function to count number ` `// of prime factors of x ` `static` `int` `countPrimeFactors(` `int` `n) ` `{ ` ` ` `if` `(n == ` `1` `) ` ` ` `return` `0` `; ` ` ` ` ` `// Count variable is ` ` ` `// incremented form every ` ` ` `// prime factor of x ` ` ` `int` `cnt = ` `0` `; ` ` ` `while` `(n % ` `2` `== ` `0` `) ` ` ` `{ ` ` ` `cnt++; ` ` ` `n = n / ` `2` `; ` ` ` `} ` ` ` ` ` `// Loop to count the number ` ` ` `// of the prime factors of ` ` ` `// the given number ` ` ` `for` `(` `int` `i = ` `3` `; i <= Math.sqrt(n); i += ` `2` `) ` ` ` `{ ` ` ` `while` `(n % i == ` `0` `) ` ` ` `{ ` ` ` `cnt++; ` ` ` `n = n / i; ` ` ` `} ` ` ` `} ` ` ` ` ` `if` `(n > ` `2` `) ` ` ` `cnt++; ` ` ` `return` `cnt; ` `} ` ` ` `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `x = ` `8` `; ` ` ` `int` `prime_factor_cnt = countPrimeFactors(x); ` ` ` `System.out.print(prime_factor_cnt + ` `"\n"` `); ` `} ` `} ` ` ` `// This code is contributed by Princi Singh ` |

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## Python3

`# Python3 implementation to find the ` `# maximum count of the prime factors ` `# by the count of factors of number ` `import` `math ` ` ` `# Function to count number ` `# of prime factors of x ` `def` `countPrimeFactors(n): ` ` ` ` ` `if` `(n ` `=` `=` `1` `): ` ` ` `return` `0` ` ` ` ` `# Count variable is ` ` ` `# incremented form every ` ` ` `# prime factor of x ` ` ` `cnt ` `=` `0` ` ` ` ` `while` `(n ` `%` `2` `=` `=` `0` `): ` ` ` `cnt ` `+` `=` `1` ` ` `n ` `=` `n ` `/` `/` `2` ` ` ` ` `# Loop to count the number ` ` ` `# of the prime factors of ` ` ` `# the given number ` ` ` `for` `i ` `in` `range` `(` `3` `, ` `int` `(math.sqrt(n)) ` `+` `1` `, ` `2` `): ` ` ` `while` `(n ` `%` `i ` `=` `=` `0` `): ` ` ` `cnt ` `+` `=` `1` ` ` `n ` `=` `n ` `/` `/` `i ` ` ` ` ` `if` `(n > ` `2` `): ` ` ` `cnt ` `+` `=` `1` ` ` ` ` `return` `cnt ` ` ` `# Driver Code ` `x ` `=` `8` `prime_factor_cnt ` `=` `countPrimeFactors(x) ` ` ` `print` `(prime_factor_cnt) ` ` ` `# This code is contributed by ShubhamCoder ` |

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## C#

`// C# implmentation to find the ` `// maximum count of the prime factors ` `// by the count of factors of number ` `using` `System; ` ` ` `class` `GFG{ ` ` ` `// Function to count number ` `// of prime factors of x ` `static` `int` `countPrimeFactors(` `int` `n) ` `{ ` ` ` `if` `(n == 1) ` ` ` `return` `0; ` ` ` ` ` `// Count variable is ` ` ` `// incremented form every ` ` ` `// prime factor of x ` ` ` `int` `cnt = 0; ` ` ` `while` `(n % 2 == 0) ` ` ` `{ ` ` ` `cnt++; ` ` ` `n = n / 2; ` ` ` `} ` ` ` ` ` `// Loop to count the number ` ` ` `// of the prime factors of ` ` ` `// the given number ` ` ` `for` `(` `int` `i = 3; ` ` ` `i <= Math.Sqrt(n); i += 2) ` ` ` `{ ` ` ` `while` `(n % i == 0) ` ` ` `{ ` ` ` `cnt++; ` ` ` `n = n / i; ` ` ` `} ` ` ` `} ` ` ` ` ` `if` `(n > 2) ` ` ` `cnt++; ` ` ` ` ` `return` `cnt; ` `} ` ` ` `// Driver Code ` `static` `public` `void` `Main() ` `{ ` ` ` `int` `x = 8; ` ` ` `int` `prime_factor_cnt = countPrimeFactors(x); ` ` ` `Console.Write(prime_factor_cnt); ` `} ` `} ` ` ` `// This code is contributed by ShubhamCoder ` |

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**Output:**

3

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