To find whether a number is odd or even?

Answer 1

The following simply uses the definition of "odd" and "even". Do an integer division by 2. If it is divisible (no remainder), it is even. Else, it is odd. Here is some sample code in Java:

System.out.println(myNumber % 2 == 0 ? "It is even" : "It is odd");

Answer 2

The way to do this is to test on the modulo (remainder) of dividing the number by 2.

For example:

4/2 = 2, with 0 remainder. So, 4 modulo 2 = 0. This is for an even number.
5/2 = 2, with 1 remainder. So, 5 modulo 2 = 1. This is for an odd number.

The modulo for any integer divided by 2 is either 1 or 0. All odd numbers will return a 1 modulo, and all even numbers return a 0 modulo.

So, your test becomes:
If the modulo of the input number is 1, then the input number is odd. Otherwise it is even.

The actual code to do this varies depending on which programming or scripting language you are going to use. For example, in Perl:

print ($ARGV[0] % 2) ? "Odd\n" : "Even\n";

Answer 3

To find out if number is Odd or Even in C Programming language you could use % operator.

It returns reminder left after division of two passed numbers.

For example:

0 % 2 = 0

4 % 3 = 1

5 % 1 = 0

To check if number is Odd or Even you need to divide it by 2.

Here is small program:

#include

int main() {

int num;

printf("Enter number: ");

scanf("%d", &num);

if (num % 2 == 0) {

printf("Number is Even.\n");

} else {

printf("Number is Odd.\n");

}

return 0;

}

Testing:

Enter number: 2

Number is Even.

Enter number: 3

Number is Odd.

Answer 4

// To check if a number is even, check to see if it is evenly divisible by 2.

// If you have no remainder, then it's even. Otherwise it's odd.

public static boolean isEven(int n) {

return (n % 2) == 0;

}

Answer 5

#include<stdoio.h>

#include<conio.h>

void main()

{

int a;

printf("Enter the number");

scanf("%d",&a);

if(a%2==0)

{

print("number is even");

}

else

{

printf("number id odd");

}

getch();

}