Oh good old-fashioned C.
void main()
{
int variable_name = [Any number goes here];
if (variable_name % 2 == 0)
{
printf("%d is even.", variable_name);
} else
{
printf("%d is odd.", variable_name);
}
}
I think I've helped enough, so it's up to you to learn how to get input from the user, if that's what you're working on.
/*
Title: even.c
A simple C program to find if the given number is even or not.
*/
#include
int main()
{
int number;
/* prompt the user for a number */
printf("\nNumber: ");
scanf("%d" , &number);
/*
Check if the number is even or not. A number is even if it is divisible by 2.
The remainder is zero.
*/
if(number % 2 == 0)
printf("The number %d is even\n" , number);
else
printf("The number %d is odd\n" , number);
return 0;
}
echo "Program to check even or odd number"echo "Enter a number"read na=`expr $n % 2`if [ $a -eq 0 ] ; then #Semicolon is most important for Executing ifelse statementsecho "It is an even number"elseecho "It is an odd number"fi
find even number in array
bool is_even(long int num) { return !(num & 1); //when the number is even(divisible by two), //its least significant bit is 0 }
using system; { public static void main(string[] args) { int n; n=int.parse(console.ReadLine()); if(n%2==0) console.WriteLine("Even number"); else console.WriteLine("Odd number"); } }
To write a C program to determine if something is odd or even you need to be a programmer. To write a program in C is complicate and only done by programmers.
No, thanks.
echo "Program to check even or odd number"echo "Enter a number"read na=`expr $n % 2`if [ $a -eq 0 ] ; then #Semicolon is most important for Executing ifelse statementsecho "It is an even number"elseecho "It is an odd number"fi
No. A given number need not even be divisible by a given prime.
The pseudo code would be as follows (you figure out the syntax) 1) Prompt the user to enter a number 2) If entered number is alpha, quit program after displaying message that the user ended the program. 3) Otherwise, find Modulo 2 of the entered number. This is a fancy way of saying "find the remainder when the number is divided by 2) 4) If Modulo 2 is zero, the number is even, otherwise odd 5) Display message showing if the entered number was Even or Odd 6) Branch back to step 1
#include<stdio.h> #include<conio.h> void main() { int n; printf("ENTER NO FOR CHECK........ "); scanf("%d", &n); if(n%2==0) printf("Given No is EVEN"); else printf("Given No is ODD"); getch(); }
int main () { int x; printf ("Enter the value of x\n"); scanf ("%d",&x); if (0 == x%2) printf ("Number is even\n") else printf ("Number is odd\n"); }
The only even prime number is 2.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2.
Assuming you know that your number is a perfect square, the square root of an even number is even, and the square root of an odd number is odd.
If it is even
find even number in array
If it is even, it is divisible by two evenly.