Check whether a given number is even or odd

Given a number, check whether it is even or odd.

Examples :

Input: 2 
Output: even

Input: 5
Output: odd

One simple solution is to find the remainder after dividing by 2. 

C++

// A simple C++ program to
// check for even or odd
#include <iostream>
using namespace std;

// Returns true if n is
// even, else odd
bool isEven(int n) { return (n % 2 == 0); }

// Driver code
int main()
{
    int n = 101;
    isEven(n) ? cout << "Even" : cout << "Odd";

    return 0;
}

Java

// Java program  to
// check for even or odd

class GFG
{
    // Returns true if n is even, else odd
    public static boolean isEven(int n)
    {
        return (n % 2 == 0);
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 101;
        if(isEven(n) == true)
            System.out.print("Even");
        else
            System.out.print("Odd");
    }
}

// This code is contributed by rishabh_jain

Python3

# A simple Python3 code
# to check for even or odd

# Returns true if n is even, else odd
def isEven(n):
    return (n % 2 == 0)

# Driver code
n = 101
print("Even" if isEven(n) else "Odd")

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program to
// check for even or odd
using System;

class GFG
{
    // Returns true if n is even, else odd
    public static bool isEven(int n)
    {
        return (n % 2 == 0);
    }

    // Driver code
    public static void Main()
    {
        int n = 101;
        if(isEven(n) == true)
            Console.WriteLine("Even");
        else
            Console.WriteLine("Odd");
    }
}

// This code is contributed by vt_m

PHP

<?php
// A simple PHP program to
// check for even or odd

// Returns true if n is
// even, else odd
function isEven($n)
{
    return ($n % 2 == 0);
}

// Driver code
$n = 101;
if(isEven != true)
    echo "Even";
    else
    echo "Odd";

// This code is contributed by Ajit
?>

Javascript

<script>

// A simple Javascript program to
// check for even or odd


// Returns true if n is
// even, else odd
function isEven(n) { return (n % 2 == 0); }

// Driver code

    let n = 101;
    isEven(n) ? document.write("Even") :document.write("Odd");


// This code is contributed by Mayank Tyagi

</script>

Output :

Odd

Time Complexity: O(1)

Auxiliary Space: O(1)
A better solution is to use bitwise operators. We need to check whether last bit is 1 or not. If last bit is 1 then number is odd, otherwise always even.
Explanation: 

 input : 5    // odd
   00000101              
 & 00000001                
--------------                
   00000001       
--------------

input : 8     //even
   00001000              
 & 00000001                 
--------------               
   00000000        
--------------

Below is the implementation of the idea. 

C++

// A simple C++ program to
// check for even or odd
#include <iostream>
using namespace std;

// Returns true if n is
// even, else odd
bool isEven(int n)
{

// n & 1 is 1, then
// odd, else even
return (!(n & 1));
}

// Driver code
int main()
{
int n = 101;
isEven(n)? cout << "Even" :
           cout << "Odd";

return 0;
}

C

#include <stdio.h>
#include <math.h>

int main(){
  int n = 101;
  if (n%2==0){
    printf("Even");
  }
  else{
    printf("Odd");
  }
 return 0;
}

Java

// Java program to
// check for even or odd

class GFG
{
    // Returns true if n
    // is even, else odd
    public static boolean isEven(int n)
    {
        if((n & 1) == 0)
            return true;
        else
            return false;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 101;
        if(isEven(n) == true)
            System.out.print("Even");
        else
            System.out.print("Odd");
    }
}

// This code is contributed by rishabh_jain

Python3

# A Python3 code program
# to check for even or odd

# Returns true if n is even, else odd
def isEven(n):

    # n&1 is 1, then odd, else even
    return (not(n & 1))

# Driver code
n = 101;
print("Even" if isEven(n) else "Odd")

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program  to
// check for even or odd
using System;

class GFG
{
    // Returns true if n
    // is even, else odd
    public static bool isEven(int n)
    {
        if((n & 1) == 0)
            return true;
        else
            return false;
    }

    // Driver code
    public static void Main()
    {
        int n = 101;
        if(isEven(n) == true)
            Console.WriteLine("Even");
        else
            Console.WriteLine("Odd");
    }
}

// This code is contributed by vt_m.

PHP

<?php
// A simple PHP program to
// check for even or odd

// Returns true if n is
// even, else odd
function isEven($n)
{
    return (!($n & 1));
}

// Driver code
$n = 101;
if(isEven($n) == true)
    echo "Even";
else
    echo "Odd";

// This code is contributed by Smitha
?>

Javascript

<script>
// A simple JavaScript program to
// check for even or odd

// Returns true if n is
// even, else odd
function isEven(n)
{

    // n & 1 is 1, then
    // odd, else even
    return (!(n & 1));
}

// Driver code
let n = 101;
isEven(n)? document.write("Even") :
        document.write("Odd");

// This code is contributed by Manoj.
</script>

Output :

Odd 

Time Complexity: O(1)

Auxiliary Space: O(1)

This article is contributed by Prabhat Raushan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.