Question

create an algorithm that will check if a number is odd or even.Display odd if the number is odd, even otherwise.

Answer 1

Answer:

#include <stdio.h>

int main() {

int num;

printf("Enter an integer: ");

scanf("%d", &num);

if(num % 2 == 0)

printf("%d is even.", num);

else

printf("%d is odd.", num);

return 0;

}

Answer 2

An even number is an integer that is exactly divisible by 2. For example: 0, 8, -24

An odd number is an integer that is not exactly divisible by 2. For example: 1, 7, -11, 15

Program to Check Even or Odd

#include <stdio.h>

int main() {

int num;

printf("Enter an integer: ");

scanf("%d", &num);

// True if num is perfectly divisible by 2

if(num % 2 == 0)

printf("%d is even.", num);

else

printf("%d is odd.", num);

return 0;

}

Output

Enter an integer: -7

-7 is odd.

In the program, the integer entered by the user is stored in the variable num.

Then, whether num is perfectly divisible by 2 or not is checked using the modulus % operator.

If the number is perfectly divisible by 2, test expression number%2 == 0 evaluates to 1 (true). This means the number is even.

However, if the test expression evaluates to 0 (false), the number is odd.

Program to Check Odd or Even Using the Ternary Operator

#include <stdio.h>

int main() {

int num;

printf("Enter an integer: ");

scanf("%d", &num);

(num % 2 == 0) ? printf("%d is even.", num) : printf("%d is odd.", num);

return 0;

}

Output

Enter an integer: 33

33 is odd.

In the above program, we have used the ternary operator ?: instead of the if...else statement.

Hope it helps...

---------- BM'S W