This is a simple program to compute the sum of first 10 even and odd numbers and display the results.It is intended for beginner level learners of C programming.

The program is written and compiled using Dev-C++ 4.9.9.2 version compiler tool. You will find following sections in this post – problem definition, flowchart, program source code and verified output to help you understand and learn C programming language. Try to practice writing, compiling and run this program on your own.

## Problem Definition

The program for computing sum of first 10 even and odd numbers follow very simple logic, given below

- Take first 20 number, N = 20 of which half are even and half are odd numbers.
- Check each number one at a time and see if it’s divisible by 2.
- If Yes, add the number to Sum_Even.
- Else, add the number to Sum_Odd.
- Display the results
- End the program

Note:- Whenever a number is divisible by 2 it gives a remainder of 0, otherwise not.

## Flowchart – Sum of Even and Odd

## Program Code – Sum of Even and Odd

#include <iostream.h>

#include <stdio.h>

#include <conio.h>

int main()

{

int sum_even,sum_odd,N,i;

//Initialize the sum of even and sum of odd numbers

sum_even = 0;

sum_odd = 0;

// Total of 20 numbers comprise of 10 odd and 10 even numbers

// But you change the value of N to anything if required

N = 20;

//Compute the sum of even and sum of odd number

for(i = 1;i<=N;i++)

{

if((i % 2) == 0)

{

sum_even = sum_even + i;

}

else

{

sum_odd = sum_odd + i;

}

}

//Print the results now

for(i=0;i<35;i++)

printf("_"); printf("\n\n");

cout <<; "Th sum of 10 evens are"

<< "\t" << sum_even << endl;

cout << "The sum of 10 odd are"

<<"\t" << sum_odd << endl;

for(i=0;i<35;i++)

printf("_"); printf("\n\n");

getch();

return 0;

}

## Output

The output of the program is given below. Remember that the program does ask for input because the value of N is already assigned. You can change that to a program asking input at run-time.