# Check if count of even divisors of N is equal to count of odd divisors

• Last Updated : 22 Mar, 2021

Given a positive integer N, the task is to check whether the count of even divisors and odd divisors of N are equal or not. If they are same then print “YES” else print “NO”.
Examples :

Input: N = 6
Output: YES
Explanation:
Number 6 has four factors:
1, 2, 3, 6,
count of even divisors = 2 (2 and 6)
count of odd divisors = 2 (1 and 3)
Input: N = 9
Output: NO
Explanation:
count of even divisors = 0
count of odd divisors = 3 (1, 3 and 9)

Naive Approach: The naive approach is to find all the divisors of the given number and count the even divisors and odd divisors and check whether they are equal or not. If they are same then print “YES”, else print “NO”.
Below is the implementation of the above approach:

## C++

 // C++ program for the above approach#include using namespace std; // Function to check if count of even// and odd divisors are equalbool divisorsSame(int n){    // To store the count of even    // factors and odd factors    int even_div = 0, odd_div = 0;     // Loop till [1, sqrt(N)]    for (int i = 1; i <= sqrt(n); i++) {        if (n % i == 0) {             // If divisors are equal            // add only one            if (n / i == i) {                 // Check for even                // divisor                if (i % 2 == 0) {                    even_div++;                }                 // Odd divisor                else {                    odd_div++;                }            }             // Check for both divisor            // i.e., i and N/i            else {                 // Check if i is odd                // or even                if (i % 2 == 0) {                    even_div++;                }                else {                    odd_div++;                }                 // Check if N/i is odd                // or even                if (n / i % 2 == 0) {                    even_div++;                }                else {                    odd_div++;                }            }        }    }     // Return true if count of even_div    // and odd_div are equals    return (even_div == odd_div);} // Driver Codeint main(){    // Given Number    int N = 6;     // Function Call    if (divisorsSame(N)) {        cout << "Yes";    }    else {        cout << "No";    }    return 0;}

## Java

 // Java code for the above programimport java.util.*; class GFG{ // Function to check if count of// even and odd divisors are equalstatic boolean divisorsSame(int n){         // To store the count of even    // factors and odd factors    int even_div = 0, odd_div = 0;     // Loop till [1, sqrt(N)]    for(int i = 1; i <= Math.sqrt(n); i++)    {       if (n % i == 0)       {                       // If divisors are equal           // add only one           if (n / i == i)           {                               // Check for even               // divisor               if (i % 2 == 0)               {                   even_div++;               }                               // Odd divisor               else               {                   odd_div++;               }           }                       // Check for both divisor           // i.e., i and N/i           else           {                               // Check if i is odd               // or even               if (i % 2 == 0)               {                   even_div++;               }               else               {                   odd_div++;               }                               // Check if N/i is odd               // or even               if (n / i % 2 == 0)               {                   even_div++;               }               else               {                   odd_div++;               }           }       }    }         // Return true if count of even_div    // and odd_div are equals    return (even_div == odd_div);} // Driver codepublic static void main(String[] args){    // Given number    int N = 6;     // Function call    if (divisorsSame(N))    {        System.out.println("Yes");    }    else    {        System.out.println("No");    }}} // This code is contributed by offbeat

## Python3

 # Python3 program for the above approachimport math # Function to check if count of even# and odd divisors are equaldef divisorsSame(n):     # To store the count of even    # factors and odd factors    even_div = 0; odd_div = 0;     # Loop till [1, sqrt(N)]    for i in range(1, int(math.sqrt(n))):             if (n % i == 0):             # If divisors are equal            # add only one            if (n // i == i):                 # Check for even                # divisor                if (i % 2 == 0):                    even_div += 1;                                 # Odd divisor                else:                    odd_div += 1;             # Check for both divisor            # i.e., i and N/i            else:                 # Check if i is odd                # or even                if (i % 2 == 0):                    even_div += 1;                                 else:                    odd_div += 1;                                 # Check if N/i is odd                # or even                if (n // (i % 2) == 0):                    even_div += 1;                                 else:                    odd_div += 1;                     # Return true if count of even_div    # and odd_div are equals    return (even_div == odd_div); # Driver Code # Given NumberN = 6; # Function Callif (divisorsSame(N) == 0):    print("Yes"); else:    print("No"); # This code is contributed by Code_Mech

## C#

 // C# code for the above programusing System;class GFG{ // Function to check if count of// even and odd divisors are equalstatic bool divisorsSame(int n){         // To store the count of even    // factors and odd factors    int even_div = 0, odd_div = 0;     // Loop till [1, sqrt(N)]    for(int i = 1; i <= Math.Sqrt(n); i++)    {        if (n % i == 0)        {                             // If divisors are equal            // add only one            if (n / i == i)            {                                     // Check for even                // divisor                if (i % 2 == 0)                {                    even_div++;                }                                     // Odd divisor                else                {                    odd_div++;                }            }                             // Check for both divisor            // i.e., i and N/i            else            {                                     // Check if i is odd                // or even                if (i % 2 == 0)                {                    even_div++;                }                else                {                    odd_div++;                }                                     // Check if N/i is odd                // or even                if (n / i % 2 == 0)                {                    even_div++;                }                else                {                    odd_div++;                }            }        }    }         // Return true if count of even_div    // and odd_div are equals    return (even_div == odd_div);} // Driver codepublic static void Main(){    // Given number    int N = 6;     // Function call    if (divisorsSame(N))    {        Console.Write("Yes");    }    else    {        Console.Write("No");    }}} // This code is contributed by Akanksha_Rai

## Javascript

 

Output:

Yes

Time Complexity: O(√N), where N is the given number
Auxiliary Space: O(1)
Efficient Approach: The idea is to observe that the numbers whose count of even and odd divisors are equals forms the following Arithmetic Progression

2, 6, 10, 14, 18, 22, …

1. The Kth term of the above series is: 2. Now we have to check whether N is a term of the above series or not by the equation:

=> => 1.
2. If the value of K calculated using the above formula is an integer, then N is the number with equal count of even and odd divisors.
3. Else N is not the number with equal count of even and odd divisors.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach#include using namespace std; // Function to check if count of even// and odd divisors are equalbool divisorsSame(int n){     // If (n-2)%4 is an integer, then    // return true else return false    return (n - 2) % 4 == 0;} // Driver Codeint main(){    // Given Number    int N = 6;     // Function Call    if (divisorsSame(N)) {        cout << "Yes";    }    else {        cout << "No";    }    return 0;}

## Java

 // Java code for the above programimport java.util.*; class GFG{     // Function to check if count of// even and odd divisors are equalstatic boolean divisorsSame(int n){         // If (n-2)%4 is an integer, then    // return true else return false    return (n - 2) % 4 == 0;} // Driver codepublic static void main(String[] args){         // Given number    int N = 6;     // Function call    if (divisorsSame(N))    {        System.out.println("Yes");    }    else    {        System.out.println("No");    }    }} // This code is contributed by offbeat

## Python3

 # Python3 program for the above approach # Function to check if count of even# and odd divisors are equaldef divisorsSame(n):     # If (n-2)%4 is an integer, then    # return true else return false    return (n - 2) % 4 == 0; # Driver Code # Given NumberN = 6; # Function Callif (divisorsSame(N)):    print("Yes");else:    print("No"); # This code is contributed by Nidhi_biet

## C#

 // C# code for the above programusing System;class GFG{     // Function to check if count of// even and odd divisors are equalstatic bool divisorsSame(int n){         // If (n-2)%4 is an integer, then    // return true else return false    return (n - 2) % 4 == 0;} // Driver codepublic static void Main(){         // Given number    int N = 6;     // Function call    if (divisorsSame(N))    {        Console.Write("Yes");    }    else    {        Console.Write("No");    }    }} // This code is contributed by Code_Mech

## Javascript

 

Output:

Yes

Time complexity: O(1)
Auxiliary Space: O(1)

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