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