We can find a number is odd or even by a simple programmain(){int a[2],i;a[0]=0; //0--means Even Numbera[1]=1; //1--means Odd numberscanf("%d",&i);printf("%d",a[i%2]);getch();}

By taking a character array its still simple.
main(){
char a[2][5]={"EVEN","ODD"};
int i;

scanf("%d",&i);
printf("%s",a[i%2]);
getch();
}

main()
(
int a,b,c;
printf("Enter the number");
scanf("%d", &a);
b=a%2;
c=b*2;
if(a=c)
printf("The number is even");
else
Printf("The number is odd");
}

This can done using switch statement

Switch(num%2)
{
Case 1:
Printf(?This is a odd number?);
Break;
Case 0:
Printf (?This is a even Number?);
}

num%2?printf("This is Odd number"):printf("This is even number");

Use ternary operatos:

void main()
{
int a;

printf(" Enter an integer number: ");
scanf("%d",&a);

((a%2)==0)?printf(" Even"):printf(" odd");

getch();
}

just make use of ternary operator:

#include<stdio.h>
#include<conio.h>

void main()
{
int a;
printf(" enter an integer number:");
scanf("%d",&a);

((k%2)==0)?printf(" %d is even",a):printf(" %d is odd",a);
getch();
}

void main()
{
int a;
printf("enter the number ");
scanf("%d",&a);
((a%2)==0))?
printf("even"):printf("odd");
gech();
}

To find Odd or Even With out using if, while or for
main()
{
int a;
scanf("%d",&a);
(a%2)?printf("Odd Number"):printf("Even Nember);
}

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