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Check if count of divisors is even or odd
• Difficulty Level : Easy
• Last Updated : 07 Feb, 2020

Given a number “n”, find its total number of divisors are even or odd.

Examples :

```Input  : n = 10
Output : Even

Input:  n = 100
Output: Odd

Input:  n = 125
Output: Even```

## We strongly recommend that you click here and practice it, before moving on to the solution.

A naive approach would be to find all the divisors and then see if the total number of divisors is even or odd.

Time complexity for such a solution would be O(sqrt(n))

## C++

 `// Naive Solution to find  ` `// if count of divisors ` `// is even or odd ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count ` `// the divisors ` `void` `countDivisors(``int` `n) ` `{ ` `    ``// Initialize count ` `    ``// of divisors ` `    ``int` `count = 0; ` ` `  `    ``// Note that this ` `    ``// loop runs till ` `    ``// square root ` `    ``for` `(``int` `i = 1; i <= ``sqrt``(n) + 1; i++)  ` `    ``{ ` `        ``if` `(n % i == 0) ` ` `  `            ``// If divisors are ` `            ``// equal increment ` `            ``// count by one ` `            ``// Otherwise increment ` `            ``// count by 2 ` `            ``count += (n / i == i) ? 1 : 2; ` `    ``} ` ` `  `    ``if` `(count % 2 == 0) ` `        ``cout << ``"Even"` `<< endl; ` `    ``else` `        ``cout << ``"Odd"` `<< endl; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``cout << ``"The count of divisor: "``; ` `    ``countDivisors(10); ` `    ``return` `0; ` `} ` ` `  `// This code is Contributed by SHUBHAMSINGH10 `

## C

 `// Naive Solution to find  ` `// if count of divisors ` `// is even or odd ` ` `  `#include ` `#include ` ` `  `// Function to count ` `// the divisors ` `void` `countDivisors(``int` `n) ` `{ ` `    ``// Initialize count ` `    ``// of divisors ` `    ``int` `count = 0; ` ` `  `    ``// Note that this ` `    ``// loop runs till ` `    ``// square root ` `    ``for` `(``int` `i = 1; i <= ``sqrt``(n) + 1; i++)  ` `    ``{ ` `        ``if` `(n % i == 0) ` ` `  `            ``// If divisors are ` `            ``// equal increment ` `            ``// count by one ` `            ``// Otherwise increment ` `            ``// count by 2 ` `            ``count += (n / i == i) ? 1 : 2; ` `    ``} ` ` `  `    ``if` `(count % 2 == 0) ` `        ``printf``(``"Even\n"``); ` ` `  `    ``else` `        ``printf``(``"Odd\n"``); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``printf``(``"The count of divisor: "``); ` `    ``countDivisors(10); ` `    ``return` `0; ` `} `

## Java

 `// Naive Solution to find if count  ` `// of divisors is even or odd ` ` `  `import` `java.io.*; ` `import` `java.math.*; ` ` `  `class` `GFG  ` `{ ` ` `  `    ``// Function to count ` `    ``// the divisors ` `    ``static` `void` `countDivisors(``int` `n) ` `    ``{ ` `        ``// Initialize count ` `        ``// of divisors ` `        ``int` `count = ``0``; ` ` `  `        ``// Note that this ` `        ``// loop runs till ` `        ``// square root ` `        ``for` `(``int` `i = ``1``; i <= Math.sqrt(n) + ``1``; i++)  ` `        ``{ ` `            ``if` `(n % i == ``0``) ` ` `  `                ``// If divisors are ` `                ``// equal increment ` `                ``// count by one ` `                ``// Otherwise increment ` `                ``// count by 2 ` `                ``count += (n / i == i) ? ``1` `: ``2``; ` `        ``} ` ` `  `        ``if` `(count % ``2` `== ``0``) ` `            ``System.out.println(``"Even"``); ` ` `  `        ``else` `            ``System.out.println(``"Odd"``); ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``System.out.print(``"The count of divisor: "``); ` `        ``countDivisors(``10``); ` `    ``} ` `} ` `// This code is contributed by Nikita Tiwari `

## Python

 `# Naive Solution to find if count   ` `# of divisors is even or odd  ` `import` `math ` ` `  `# Function to count  ` `# the divisors ` `def` `countDivisors(n) : ` `     `  `    ``# Initialize count  ` `    ``# of divisors ` `    ``count ``=` `0` ` `  `    ``# Note that this loop  ` `    ``# runs till square  ` `    ``# root ` `    ``for` `i ``in` `range``(``1``, (``int``)(math.sqrt(n)) ``+` `2``) : ` `        ``if` `(n ``%` `i ``=``=` `0``) : ` `             `  `            ``# If divisors are ` `            ``# equal, increment ` `            ``# count by one ` `            ``# Otherwise increment ` `            ``# count by 2 ` `            ``if``( n ``/``/` `i ``=``=` `i) : ` `                ``count ``=` `count ``+` `1` `            ``else` `: ` `                ``count ``=` `count ``+` `2` ` `  `    ``if` `(count ``%` `2` `=``=` `0``) : ` `        ``print``(``"Even"``) ` `    ``else` `: ` `        ``print``(``"Odd"``) ` ` `  ` `  `# Driver Code ` `print``(``"The count of divisor: "``) ` `countDivisors(``10``) ` ` `  `# This code is contributed by Nikita Tiwari `

## C#

 `// C# program using Naive ` `// Solution to find if ` `// count of divisors ` `// is even or odd ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to count ` `    ``// the divisors ` `    ``static` `void` `countDivisors(``int` `n) ` `    ``{ ` `        ``// Initialize count ` `        ``// of divisors ` `        ``int` `count = 0; ` ` `  `        ``// Note that this ` `        ``// loop runs till ` `        ``// square root ` `        ``for` `(``int` `i = 1; i <= Math.Sqrt(n) ` `                                 ``+ 1; ` `             ``i++) { ` `            ``if` `(n % i == 0) ` ` `  `                ``// If divisors are ` `                ``// equal increment ` `                ``// count by one ` `                ``// Otherwise increment ` `                ``// count by 2 ` `                ``count += (n / i == i) ? 1 : 2; ` `        ``} ` ` `  `        ``if` `(count % 2 == 0) ` `            ``Console.Write(``"Even"``); ` ` `  `        ``else` `            ``Console.Write(``"Odd"``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``Console.Write(``"The count of divisor: "``); ` `        ``countDivisors(10); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007. `

## PHP

 ` `

Output :

```The count of divisor: Even
```

Efficient Solution:
We can observe that the number of divisors is odd only in case of perfect squares. Hence the best solution would be to check if the given number is perfect square or not. If it’s a perfect square, then the number of divisors would be odd, else it’d be even.

Below is the implementation of above idea :

## C++

 `// C++ program for ` `// Efficient Solution to find ` `// if count of divisors is ` `// even or odd ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find if count ` `// of divisors is even or ` `// odd ` `void` `countDivisors(``int` `n) ` `{ ` `    ``int` `root_n = ``sqrt``(n); ` ` `  `    ``// If n is a perfect square, ` `    ``// then it has odd divisors ` `    ``if` `(root_n * root_n == n) ` `        ``printf``(``"Odd\n"``); ` `    ``else` `        ``printf``(``"Even\n"``); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``cout << ``"The count of divisors"` `         ``<< ``" of 10 is: "``; ` ` `  `    ``countDivisors(10); ` `    ``return` `0; ` `} `

## Java

 `// Java program for Efficient  ` `// Solution to find if count of  ` `// divisors is even or odd ` `import` `java.io.*; ` `import` `java.math.*; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Function to find if count ` `    ``// of divisors is even or ` `    ``// odd ` `    ``static` `void` `countDivisors(``int` `n) ` `    ``{ ` `        ``int` `root_n = (``int``)(Math.sqrt(n)); ` ` `  `        ``// If n is a perfect square, ` `        ``// then, it has odd divisors ` `        ``if` `(root_n * root_n == n) ` `            ``System.out.println(``"Odd"``); ` ` `  `        ``else` `            ``System.out.println(``"Even"``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `        ``throws` `IOException ` `    ``{ ` `        ``System.out.print(``"The count of"` `+  ` `                    ``"divisors of 10 is: "``); ` ` `  `        ``countDivisors(``10``); ` `    ``} ` `} ` ` `  `// This code is contributed by Nikita Tiwari `

## Python

 `# Python program for ` `# Efficient Solution to find ` `# find if count of divisors ` `# is even or odd ` ` `  `def` `NumOfDivisor(n): ` `    ``if` `n < ``1``: ` `        ``return` `    ``root_n ``=` `n``*``*``0.5` `     `  `    ``# If n is a perfect square, ` `    ``# then it has odd divisors ` `    ``if` `root_n``*``*``2` `=``=` `n: ` `        ``print``(``'Odd'``) ` `    ``else``: ` `        ``print``(``'Even'``) ` `         `  `# Driver code      ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``print``(``"The count of divisor"``+` `          ``"of 10 is: "``) ` `    ``NumOfDivisor(``10``) ` `     `  `# This code is contributed by Yt R     `

## C#

 `// C# program for efficient ` `// solution to find of ` `// count of divisors is ` `// even or odd ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// Function to find if ` `    ``// count of divisors ` `    ``// is even or odd ` `    ``static` `void` `countDivisors(``int` `n) ` `    ``{ ` `        ``int` `root_n = (``int``)(Math.Sqrt(n)); ` ` `  `        ``// If n is a perfect square, ` `        ``// then, it has odd divisors ` `        ``if` `(root_n * root_n == n) ` `            ``Console.WriteLine(``"Odd"``); ` ` `  `        ``else` `            ``Console.WriteLine(``"Even"``); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``Console.Write(``"The count of divisors : "``); ` ` `  `        ``countDivisors(10); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007. `

## PHP

 ` `

Output :

```The count of divisor: Even
```