C program for odd or even number





Code 1:
1. C program to check even or odd
2. C determine odd or even
3. How to check odd number in c
4. How to determine odd or even in c
5. C even odd test

#include<stdio.h>

int main(){

    int number;
  
    printf("Enter any integer: ");
    scanf("%d",&number);

    if(number % 2 ==0)
         printf("%d is even number.",number);
    else
         printf("%d is odd number.",number);
  
    return 0;

}

Sample output:
Enter any integer: 5
5 is odd number.

Code 2:
1. Display odd numbers in c
2. How to print odd numbers in c

#include<stdio.h>

int main(){

    int number;
    int min,max;
  
    printf("Enter the minimum range: ");
    scanf("%d",&min);

    printf("Enter the maximum range: ");
    scanf("%d",&max);

    printf("Odd numbers in given range are: ");
    for(number = min;number <= max; number++)

         if(number % 2 !=0)
             printf("%d ",number);
  
    return 0;

}

Sample output:
Enter the minimum range: 1
Enter the maximum range: 20
Odd numbers in given ranges are: 1 3 5 7 9 11 13 15 17 19

Code 3:
1. Even and odd numbers program in c
2. C program to find even or odd

#include<stdio.h>

int main(){

    int number;
    int min,max;
  
    printf("Enter the minimum range: ");
    scanf("%d",&min);

    printf("Enter the maximum range: ");
    scanf("%d",&max);

    printf("Odd numbers in given range are: ");
    for(number = min;number <= max; number++)

         if(number % 2 !=0)
             printf("%d ",number);

    printf("\nEven numbers in given range are: ");
    for(number = min;number <= max; number++)

         if(number % 2 ==0)
             printf("%d ",number);
  
    return 0;
}

Sample output:
Enter the minimum range: 1
Enter the maximum range: 20
Odd numbers in given ranges are: 1 3 5 7 9 11 13 15 17 19
Even numbers in given ranges are: 2 4 6 8 10 12 14 16 18 20

Code 4:
1. Sum of odd numbers in c

#include<stdio.h>

int main(){

    int number;
    int min,max;
    long sum =0;
  
    printf("Enter the minimum range: ");
    scanf("%d",&min);

    printf("Enter the maximum range: ");
    scanf("%d",&max);

    for(number = min;number <= max; number++)
         if(number % 2 !=0)
             sum = sum + number;

    printf("Sum of odd numbers in given range is: %ld",sum);
  
    return 0;

}

Sample output:
Enter the minimum range: 1
Enter the maximum range: 100
Sum of odd numbers in given range is: 2500

Code 5:
1. Sum of odd and even numbers c program

#include<stdio.h>

int main(){

    int number;
    int min,max;
    long odd_sum =0,even_sum = 0;
  
    printf("Enter the minimum range: ");
    scanf("%d",&min);

    printf("Enter the maximum range: ");
    scanf("%d",&max);

    for(number = min;number <= max; number++)
         if(number % 2 != 0)
             odd_sum = odd_sum + number;
         else
             even_sum = even_sum + number;

    printf("Sum of even numbers in given range is: %ld\n",even_sum);
    printf("Sum of odd numbers in given range is: %ld",odd_sum);
  
    return 0;

}

Sample output:
Enter the minimum range: 1
Enter the maximum range: 10
Sum of even numbers in given range is: 30
Sum of odd numbers in given range is: 25


Algorithm:

Number is called even number if it is divisible by two otherwise odd.  

Example of even numbers: 0,2,4,8,9,10 etc.
Example of odd numbers: 1, 3,5,7,9 etc.





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22 comments:

Anonymous said...

It's wasteful to force the computer to do a division ( via % ) every iteration of a loop.

for(number = min; number <= max; number++)
if(number % 2 !=0)
printf("%d ",number);

can be rewritten as

for(number = min + !(min % 2); number <= max; number += 2)
printf("%d ", number);

which will run significantly faster.

Anonymous said...

#include
void main()
{
int a,b;
printf(" odd even");
for(a=1,b=2;b<=10;b++,a++)
printf("\n%d" %d",a,b);
}

output
odd even
1 2
3 4
5 6
7 8
9 10

HAPPY said...

this can be written as
#include
int main()
{
int a,b;
printf(" odd even");
for(a=1,b=2;b<=10;b+=2,a+=2)
printf("\n%d %d",a,b);
}

alka joseph said...

thanku very much

siya rose said...

good

Anonymous said...

can u help me how you can post a program like that using synatxhighlater . can u please tell

dvrsolutions2012@gmail.com

Anonymous said...

voidmain()
inta a;
do
{
cin>>a;
if (a>0)
{
p=p+a;
cout<<p;
}
else
{
n=n+a;
cout<<n;
}
i++;
{
while (i<=100)
}
}

Anand said...

These are some more logics to check even or odd.

if((num/2)*2 == num)
then num is even
else num is odd.

or
if(num & 1)
printf("odd");
else
printf("even");

Unknown said...

can i get the program to find weather the no is even odd without using nested if and else????

Unknown said...

Or

Anonymous said...

such a really good learning site or "c"
.
.
#aBmaitra <3

sys101jimi said...

Good answer. anding with one uses the characteristics of binary.

sys101jimi said...

&

sys101jimi said...

And operation. "Or off, And in"

sys101jimi said...

And

Unknown said...

Thanks Brother
warm regard from Indonesia

fedillah koh said...

Using c language, the input and output are 0 and 1. Its output is identical to the input in the even position but inverted in the odd position. For example, input 0000111 it output 1010010. How to do this ?

আনমনে নিখিল said...

Write a program that:
Takes integer inputs until 0 is given,
1. Calculate the sum of all odd numbers,
2. Count the number of even numbers

Sample Input: 2 4 5 11 12 6 7 8 13 1 2 0
Sample Output: Total of all odd numbers is= 37
Number of even number = 6

*** Use while loop for this program. (This program can be done by both for loop and while loop. But you are instructed to do this with while loop).

Unknown said...

thanks...

Unknown said...

#include

int main()
{
int n;

printf("Enter an integer\n");
scanf("%d", &n);

if (n & 1 == 1)
printf("Odd\n");
else
printf("Even\n");

return 0;
}

can anybody solved this logic how we get odd or even value?

Unknown said...

how its works

Modder said...

Any integer number is represented in binary form.

An even number's LSB will always be 0.
An odd number will have the LSB, always 1.

For example:
decimal 5 --> binary 0x0101
decimal 4 --> binary 0x0100

You can check with any number to understand this.

The LSB will tell you if the number is even or odd.