Check whether a given number is even or odd
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- Difficulty Level : Basic
- Last Updated : 05 Jul, 2022
Given a number, check whether it is even or odd.

Examples :
Input: 2 Output: even Input: 5 Output: odd
Recommended Practice
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 oddbool isEven(int n) { return (n % 2 == 0); }// Driver codeint main(){ int n = 101; isEven(n) ? cout << "Even" : cout << "Odd"; return 0;} |
Java
// Java program to// check for even or oddclass 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 odddef isEven(n): return (n % 2 == 0) # Driver coden = 101print("Even" if isEven(n) else "Odd")# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program to// check for even or oddusing 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 oddfunction 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 oddfunction 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 oddbool isEven(int n){ // n & 1 is 1, then// odd, else evenreturn (!(n & 1));}// Driver codeint 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 oddclass 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 odddef isEven(n): # n&1 is 1, then odd, else even return (not(n & 1)) # Driver coden = 101;print("Even" if isEven(n) else "Odd")# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# program to// check for even or oddusing 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 oddfunction 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 oddfunction isEven(n){ // n & 1 is 1, then // odd, else even return (!(n & 1));}// Driver codelet 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)
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