# Check if a number has prime count of divisors

• Difficulty Level : Basic
• Last Updated : 22 Jun, 2022

Given an integer N, the task is to check if the count of divisors of N is prime or not.

Examples:

Input: N = 13
Output: Yes
The divisor count is 2 (1 and 13) which is prime.

Input: N = 8
Output: No
The divisors are 1, 2, 4 and 8.

Approach: Please read this article to find the count of divisors of a number. So find the maximum value of i for every prime divisor p such that N % (pi) = 0. So the count of divisors gets multiplied by (i + 1). The count of divisors will be (i1 + 1) * (i2 + 1) * … * (ik + 1).
It can now be seen that there can only be one prime divisor for the maximum i and if N % pi = 0 then (i + 1) should be prime. The primality can be checked in sqrt(n) time and the prime factors can also be found in sqrt(n) time. So the overall time complexity will be O(sqrt(n)).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function that returns true` `// if n is prime` `bool` `Prime(``int` `n)` `{` `    ``// There is no prime` `    ``// less than 2` `    ``if` `(n < 2)` `        ``return` `false``;`   `    ``// Run a loop from 2 to sqrt(n)` `    ``for` `(``int` `i = 2; i <= ``sqrt``(n); i++)`   `        ``// If there is any factor` `        ``if` `(n % i == 0)` `            ``return` `false``;`   `    ``return` `true``;` `}`   `// Function that returns true if n` `// has a prime count of divisors` `bool` `primeCountDivisors(``int` `n)` `{` `    ``if` `(n < 2)` `        ``return` `false``;`   `    ``// Find the prime factors` `    ``for` `(``int` `i = 2; i <= ``sqrt``(n); i++)` `        ``if` `(n % i == 0) {`   `            ``// Find the maximum value of i for every` `            ``// prime divisor p such that n % (p^i) == 0` `            ``long` `a = n, c = 0;` `            ``while` `(a % i == 0) {` `                ``a /= i;` `                ``c++;` `            ``}`   `            ``// If c+1 is a prime number and a = 1` `            ``if` `(a == 1 && Prime(c + 1))` `                ``return` `true``;`   `            ``// The number cannot have two factors` `            ``// to have count of divisors prime` `            ``else` `                ``return` `false``;` `        ``}`   `    ``// Else the number is prime so` `    ``// it has only two divisors` `    ``return` `true``;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 13;`   `    ``if` `(primeCountDivisors(n))` `        ``cout << ``"Yes"``;` `    ``else` `        ``cout << ``"No"``;`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach ` `class` `GFG` `{` `    `  `    ``// Function that returns true ` `    ``// if n is prime ` `    ``static` `boolean` `Prime(``int` `n) ` `    ``{ ` `        ``// There is no prime ` `        ``// less than 2 ` `        ``if` `(n < ``2``) ` `            ``return` `false``; ` `    `  `        ``// Run a loop from 2 to sqrt(n) ` `        ``for` `(``int` `i = ``2``; i <= (``int``)Math.sqrt(n); i++) ` `    `  `            ``// If there is any factor ` `            ``if` `(n % i == ``0``) ` `                ``return` `false``; ` `        ``return` `true``; ` `    ``} ` `    `  `    ``// Function that returns true if n ` `    ``// has a prime count of divisors ` `    ``static` `boolean` `primeCountDivisors(``int` `n) ` `    ``{ ` `        ``if` `(n < ``2``) ` `            ``return` `false``; ` `    `  `        ``// Find the prime factors ` `        ``for` `(``int` `i = ``2``; i <= (``int``)Math.sqrt(n); i++) ` `            ``if` `(n % i == ``0``) ` `            ``{ ` `    `  `                ``// Find the maximum value of i for every ` `                ``// prime divisor p such that n % (p^i) == 0 ` `                ``long` `a = n, c = ``0``; ` `                ``while` `(a % i == ``0``)` `                ``{ ` `                    ``a /= i; ` `                    ``c++; ` `                ``} ` `    `  `                ``// If c+1 is a prime number and a = 1 ` `                ``if` `(a == ``1` `&& Prime((``int``)c + ``1``) == ``true``) ` `                    ``return` `true``; ` `    `  `                ``// The number cannot have two factors ` `                ``// to have count of divisors prime ` `                ``else` `                    ``return` `false``; ` `            ``} ` `    `  `        ``// Else the number is prime so ` `        ``// it has only two divisors ` `        ``return` `true``; ` `    ``} ` `    `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args)` `    ``{ ` `        ``int` `n = ``13``; ` `    `  `        ``if` `(primeCountDivisors(n)) ` `            ``System.out.println(``"Yes"``); ` `        ``else` `            ``System.out.println(``"No"``); ` `    ``} ` `}`   `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation of the approach ` `from` `math ``import` `sqrt`   `# Function that returns true ` `# if n is prime ` `def` `Prime(n) : `   `    ``# There is no prime ` `    ``# less than 2 ` `    ``if` `(n < ``2``) :` `        ``return` `False``; `   `    ``# Run a loop from 2 to sqrt(n) ` `    ``for` `i ``in` `range``(``2``, ``int``(sqrt(n)) ``+` `1``) :`   `        ``# If there is any factor ` `        ``if` `(n ``%` `i ``=``=` `0``) :` `            ``return` `False``; `   `    ``return` `True``; `   `# Function that returns true if n ` `# has a prime count of divisors ` `def` `primeCountDivisors(n) : `   `    ``if` `(n < ``2``) :` `        ``return` `False``; `   `    ``# Find the prime factors ` `    ``for` `i ``in` `range``(``2``, ``int``(sqrt(n)) ``+` `1``) :` `        ``if` `(n ``%` `i ``=``=` `0``) :`   `            ``# Find the maximum value of i for every ` `            ``# prime divisor p such that n % (p^i) == 0 ` `            ``a ``=` `n; c ``=` `0``; ` `            ``while` `(a ``%` `i ``=``=` `0``) :` `                ``a ``/``/``=` `i; ` `                ``c ``+``=` `1``; `   `            ``# If c + 1 is a prime number and a = 1 ` `            ``if` `(a ``=``=` `1` `and` `Prime(c ``+` `1``)) :` `                ``return` `True``; `   `            ``# The number cannot have two factors ` `            ``# to have count of divisors prime ` `            ``else` `:` `                ``return` `False``; ` `        `  `    ``# Else the number is prime so ` `    ``# it has only two divisors ` `    ``return` `True``; `   `# Driver code ` `if` `__name__ ``=``=` `"__main__"` `: `   `    ``n ``=` `13``; `   `    ``if` `(primeCountDivisors(n)) :` `        ``print``(``"Yes"``); ` `    ``else` `:` `        ``print``(``"No"``); `   `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation of the approach ` `using` `System;`   `class` `GFG ` `{ ` `    `  `    ``// Function that returns true ` `    ``// if n is prime ` `    ``static` `bool` `Prime(``int` `n) ` `    ``{ ` `        `  `        ``// There is no prime ` `        ``// less than 2 ` `        ``if` `(n < 2) ` `            ``return` `false``; ` `    `  `        ``// Run a loop from 2 to sqrt(n) ` `        ``for` `(``int` `i = 2; i <= (``int``)Math.Sqrt(n); i++) ` `    `  `            ``// If there is any factor ` `            ``if` `(n % i == 0) ` `                ``return` `false``; ` `        ``return` `true``; ` `    ``} ` `    `  `    ``// Function that returns true if n ` `    ``// has a prime count of divisors ` `    ``static` `bool` `primeCountDivisors(``int` `n) ` `    ``{ ` `        ``if` `(n < 2) ` `            ``return` `false``; ` `    `  `        ``// Find the prime factors ` `        ``for` `(``int` `i = 2; i <= (``int``)Math.Sqrt(n); i++) ` `            ``if` `(n % i == 0) ` `            ``{ ` `    `  `                ``// Find the maximum value of i for every ` `                ``// prime divisor p such that n % (p^i) == 0 ` `                ``long` `a = n, c = 0; ` `                ``while` `(a % i == 0) ` `                ``{ ` `                    ``a /= i; ` `                    ``c++; ` `                ``} ` `    `  `                ``// If c+1 is a prime number and a = 1 ` `                ``if` `(a == 1 && Prime((``int``)c + 1) == ``true``) ` `                    ``return` `true``; ` `    `  `                ``// The number cannot have two factors ` `                ``// to have count of divisors prime ` `                ``else` `                    ``return` `false``; ` `            ``} ` `    `  `        ``// Else the number is prime so ` `        ``// it has only two divisors ` `        ``return` `true``; ` `    ``} ` `    `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `n = 13; ` `    `  `        ``if` `(primeCountDivisors(n)) ` `            ``Console.WriteLine(``"Yes"``); ` `        ``else` `            ``Console.WriteLine(``"No"``); ` `    ``} ` `} `   `// This code is contributed by AnkitRai01 `

## Javascript

 ``

Output:

`Yes`

Time Complexity: O(sqrt(n)), as we are using a loop to traverse sqrt (n) times. Where n is the integer given as input.

Auxiliary Space: O(1), as we are not using any extra space.

My Personal Notes arrow_drop_up
Related Articles