#include<stdio.h>
int main()
{
	int n=10000;
	char *s[2]={"Even","Odd"};
	printf("%s",s[n&1]);
	return 0;
}

A number anded with the lower number that is n & (n - 1) = 
0 then it is even if it is anything else it is odd

odd_even (int n)

{
   if (!(n & (n -1))
      number is odd
   else
      number is even

}

But its much efficient just to find whether the last bit is
0 or 1

the first answer is excellent .... superb..... this is what
i expected........ marvalous.......... congrats



thank u

#include<stdio.h>
main()
{
 int n;
 char *p[]={"Even","odd"};
 Printf("Enter the number");
 scanf("%d",&n);
 n=n%2;
 printf("The value is %s",a[n]);

}

#include<stdio.h>
main()
{
 int n;
 string p[2]={"Even","odd"};
 Printf("Enter the number");
 scanf("%d",&n);
 n=n%2;
 printf("The value is %s",p[n]);

}

THE LAST two answers posted by two folks are correct but the
declarations have been made wrong...... we cant make use of
1D array here , if so only 'e' or 'o' only will get
printed.... but that is not our aim... so correct
declaration is using a 2D array.....
char a[][6]={{"even"},{"odd"}};

and also it is not the must to make use of array of pointers
concept...........

thank u

#include<stdio.h>
#include<conio.h>
void main()
{
int x;
if(x%2==0)
printf("even");
else
printf("odd");
getch();
}

mind you sir == is an relational operator.................



thank u

#include<stdio.h>
main()
{
 int n;
 string s[2]={"Even","odd"};
 Printf("Enter the number");
 scanf("%d",&n);
 n=n%2;
 printf("The value is %s",s[n]);

}

ya. the first answer has impressed me.

#include<stdio.h>
main()
{
 int n;
 string p[2]={"Even","odd"};
 Printf("Enter the number");
 scanf("%d",&n);
 n=n%2;
 printf("The value is %s",p[n]);

}

#include<stdio.h>
#include<conio.h>
main()
{
 int n;
 char *p[]={"Even","odd"};
 clrscr();
 printf("Enter the number");
 scanf("%d",&n);
 n=n%2;
 printf("The value is %s",p[n]);
 getch();

}

int oddeven(int n)
{
if(n&1)
return 1; //odd

else
return 0; //even
}

#include<stdio.h>
void main()
{
int p;
printf("number:");
scanf("%d",&p);
while(p%2)
{
printf("odd");
}
printf("even");
}