The way to do this is to test on the modulo (remainder) of dividing the number by 2.
For example:
4/2 = 2, with 0 remainder. So, 4 modulo 2 = 0. This is for an even number.
5/2 = 2, with 1 remainder. So, 5 modulo 2 = 1. This is for an odd number.
The modulo for any integer divided by 2 is either 1 or 0. All odd numbers will return a 1 modulo, and all even numbers return a 0 modulo.
So, your test becomes:
If the modulo of the input number is 1, then the input number is odd. Otherwise it is even.
The actual code to do this varies depending on which programming or scripting language you are going to use. For example, in Perl:
print ($ARGV[0] % 2) ? "Odd\n" : "Even\n";
To find out if number is Odd or Even in C Programming language you could use % operator.
It returns reminder left after division of two passed numbers.
For example:
0 % 2 = 0
4 % 3 = 1
5 % 1 = 0
To check if number is Odd or Even you need to divide it by 2.
Here is small program:
#include
int main() {
int num;
printf("Enter number: ");
scanf("%d", &num);
if (num % 2 == 0) {
printf("Number is Even.\n");
} else {
printf("Number is Odd.\n");
}
return 0;
}
Testing:
Enter number: 2
Number is Even.
Enter number: 3
Number is Odd.
first we write start and then read number and after that check the number is totaly divide by 2 or not if number is totally divide by 2 then number is even else number is odd.
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
It should not be even (except for 2), and must not be dividable with any odd number between 3 and its square-root.
printf(\n "ENTER THE NUMBER\t"); scanf{"%d",&a); while(a!=0); { r=a%2; if(r==0) printf("\n\n THE NUMBER IS EVEN \t"); else printf("\n\n THE NUMBER IS ODD \t"); printf ("\n ENTER THE NUMBER \t"); scanf("%d",&a); } getch(); }
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"); } }
The procedure is the same whether the number is even or odd. There is no separate procedure for odd numbers.
No. All number ending with 0,2,4,6,8 is even and numbers with 1,3,5,7,9 endings are odd. Since 50 has an ending of 0, it is even. Another way to find whether or not a number is even or odd, you just have to divide that number by 2. If the answer is a whole number, it is even.
It is a negative number. The answer is the same whether there is an odd or even number of addends.
whenever you add an even number the answer will be the same (odd/even) as the other number
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2.
An even number can be divided by 2 evenly. An odd number will have a remainder of 1 when divided by 2.
It is a negative number. The answer is the same whether there is an odd or even number of addends.
If the factorization includes the number 2, it's even. If not, it's odd.
first we write start and then read number and after that check the number is totaly divide by 2 or not if number is totally divide by 2 then number is even else number is odd.
The sign of a rational number does not depend on whether it is odd or even.
It's just one of those laws of maths. It can be used as an indication to whether the answer could be right. It's similar to even + even = even Even + odd = odd Odd + odd = even
A number is even when you can divide it by 2. Ex: 44/2=22 (even) 18436/2=8218 (even) 165/2=82.5 (odd)