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C Program To Find Sum of All Even Numbers Between Two Integers, using While loop

Lets write a C program to find sum of all the even numbers between range or between 2 integers input by the user.

Even Number: An even number is an integer that is exactly divisible by 2.

For Example: 10 % 2 == 0. When we divide 10 by 2, it give a reminder of 0. So number 10 is an even number.

Note: In this C program we ask the user to input start and end value. We assume that the user enters bigger value for variable end and smaller value for variable start. i.e., start < end

If user enters start = 10 and end = 20. C program finds all the even numbers between 10 and 20, including 10 and 20. So the even numbers are 10, 12, 14, 16, 18, 20. We add all these even numbers and output the sum to the console window. i.e., 10 + 12 + 14 + 16 + 18 + 20 = 90. We out put the value 90 as result.

Video Tutorial: C Program To Find Sum of All Even Numbers Between Range, using While loop

[youtube https://www.youtube.com/watch?v=QX3DExBMohQ]

YouTube Link: https://www.youtube.com/watch?v=QX3DExBMohQ [Watch the Video In Full Screen.]

Logic To Find Sum of All Even Numbers Between Two Integers, using While loop

Step 1: We ask the user to enter start and end value.

Step 2: We iterate through the while loop until value of start is less than or equal to value of variable end. Inside while loop we keep incrementing the value of variable start by one for each iteration.

Step 3: For every iteration we check if value present in variable start is a even number. i.e., start % 2 == 0. If this condition is true, then we add the value present in variable start to the previous value of variable sum.

Step 4: Once the control exits while loop, we print the value present in variable sum – which has the sum of all the even numbers between the range entered by the user.

Source Code: C Program To Find Sum of All Even Numbers Between Two Integers, using While loop

#include<stdio.h>
int main()
{
    int start, end, sum = 0;

    printf("Enter start and end value\n");
    scanf("%d%d", &start, &end);

    printf("\nSum of even no's from %d to %d is ", start, end);
    while(start <= end)
    {
        if(start % 2 == 0)
        {
            sum = sum + start;
        }
        start++;
    }
    printf("%d\n", sum);

    return 0;
}

Output 1:
Enter start and end value
10
20

Sum of even no’s from 10 to 20 is 90

Output 2:
Enter start and end value
25
50

Sum of even no’s from 25 to 50 is 494

For list of all c programming interviews / viva question and answers visit: C Programming Interview / Viva Q&A List

For full C programming language free video tutorial list visit:C Programming: Beginner To Advance To Expert

C Program To Find Sum of All Even Numbers From 1 To N, using While loop

Lets write a C program to find sum of all the even numbers from 1 to N, using while loop.

Even Number: An even number is an integer that is exactly divisible by 2.

For Example: 8 % 2 == 0. When we divide 8 by 2, it give a reminder of 0. So number 8 is an even number.

If user enters num = 5. Even numbers between 1 to 5 are 2, 4. So we add 2 and 4. i.e., 2 + 4 = 6. We display 6 to the console window as result.

Video Tutorial: C Program To Find Sum of All Even Numbers From 1 To N, using While loop

[youtube https://www.youtube.com/watch?v=EZeusHqc0Rc]

YouTube Link: https://www.youtube.com/watch?v=EZeusHqc0Rc [Watch the Video In Full Screen.]

Logic To Find Sum of All Even Numbers From 1 To N, using While loop

Since we are checking for even numbers from 1 to user entered number, we assign value of variable count to 1. While loop keeps iterating until value of count is less than or equal to value of user input number. For each iteration of while loop we increment the value of count by 1.

Inside while loop we check for the condition, count%2 == 0. If it’s true, then we add the value present inside variable count to previous value present in variable sum.

After the control exits while loop we display the value present in variable sum – which has the sum of all the even numbers from 1 to user entered number.

Source Code: C Program To Find Sum of All Even Numbers From 1 To N, using While loop

#include<stdio.h>

int main()
{
    int num, count = 1, sum = 0;

    printf("Enter the limit\n");
    scanf("%d", &num);

    while(count <= num)
    {
        if(count%2 == 0)
        {
            sum = sum + count;
        }
        count++;
    }
    printf("Sum of Even numbers from 1 to %d is %d\n", num, sum);
    return 0;
}

Output 1:
Enter the limit
5
Sum of Even numbers from 1 to 5 is 6

Output 2:
Enter the limit
20
Sum of Even numbers from 1 to 20 is 110

For list of all c programming interviews / viva question and answers visit: C Programming Interview / Viva Q&A List

For full C programming language free video tutorial list visit:C Programming: Beginner To Advance To Expert

C Program To Find Sum of All Odd Numbers From 1 To N, using While loop

Lets write a C program to find sum of all the odd numbers from 1 to N, using while loop.

Odd Number: An odd number is an integer that is not exactly divisible by 2.

For Example: 3 % 2 != 0. When we divide 3 by 2, it doesn’t give a reminder of 0. So number 3 is odd number.

Video Tutorial: C Program To Find Sum of All Odd Numbers From 1 To N, using While loop

[youtube https://www.youtube.com/watch?v=y4Sy9xo-pFU]

YouTube Link: https://www.youtube.com/watch?v=y4Sy9xo-pFU [Watch the Video In Full Screen.]

Logic To Find Sum of All Odd Numbers From 1 To N, using While loop

Since we are checking for odd numbers from 1 to user entered number, we assign value of variable count to 1. While loop keeps iterating until value of count is less than or equal to value of user input number. For each iteration of while loop we increment the value of count by 1.

Inside while loop we check for the condition, count%2 != 0. If it’s true, then we add the value present inside variable count to previous value present in variable sum.

After the control exits while loop we display the value present in variable sum – which has the sum of all the odd numbers from 1 to user entered number.

Source Code: C Program To Find Sum of All Odd Numbers From 1 To N, using While loop

#include<stdio.h>

int main()
{
    int num, count = 1, sum = 0;

    printf("Enter a integer number\n");
    scanf("%d", &num);

    while(count <= num)
    {
        if(count%2 != 0)
        {
            sum = sum + count;
        }
        count++;
    }

    printf("Sum of ODD integer number is %d\n", sum);

    return 0;
}

Output 1:
Enter a integer number
5
Sum of ODD integer number is 9

Output 2:
Enter a integer number
20
Sum of ODD integer number is 100

For list of all c programming interviews / viva question and answers visit: C Programming Interview / Viva Q&A List

For full C programming language free video tutorial list visit:C Programming: Beginner To Advance To Expert

C Program To Find Prime Numbers Between Two Intervals, using While Loop

Lets write a C program to find and print/display all the prime numbers between 2 integer values input by the user, using nested while loop.

Prime Number: Any natural number which is greater than 1 and has only two factors i.e., 1 and the number itself is called a prime number.

Video Tutorial: C Program To Find Prime Numbers Between Range, using While Loop

[youtube https://www.youtube.com/watch?v=7JgeqZz9uhw]

YouTube Link: https://www.youtube.com/watch?v=7JgeqZz9uhw [Watch the Video In Full Screen.]

Inner While loop Logic

All the numbers are perfectly divisible by number 1, so we initialize the variable count to 2, instead of 1. So our c program starts checking for divisibility from number 2.

While loop Logic

Outer while loop selects a number for each iteration and stores inside variable start. Inner while loop checks if the selected value(present in variable start) is prime or not. If its a prime number then the variable prime will have value of 1 or else it’ll have value 0 inside it(after completion of inner while loop iteration). If the value present in variable prime is 1, then we print the value present in variable start on to the console window.

Source Code: C Program To Find Prime Numbers Between Two Intervals, using While Loop

#include < stdio.h >
#include < math.h >

int main()
{
    int start, end, count, prime, inum;

    printf("Enter start and end value\n");
    scanf("%d%d", &start, &end);

    printf("\n\nPrime Numbers from %d to %d are:\n", start, end);
    while(start <= end)
    {
        inum  = sqrt(start);
        count = 2;
        prime = 1;

        while(count <= inum)
        {
            if(start%count == 0)
            {
                prime = 0;
                break;
            }
            count++;
        }

        if(prime) printf("%d, ", start);
        start++;
    }
    printf("\n\n");

    return 0;
}

Output 1:
Enter start and end value
10
50

Prime Numbers from 10 to 50 are:
11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,

Output 2:
Enter start and end value
50
100

Prime Numbers from 50 to 100 are:
53, 59, 61, 67, 71, 73, 79, 83, 89, 97,

Output 3:
Enter start and end value
100
200

Prime Numbers from 100 to 200 are:
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199,

Output 4:
Enter start and end value
25
50

Prime Numbers from 25 to 50 are:
29, 31, 37, 41, 43, 47,

Output 5:
Enter start and end value
75
150

Prime Numbers from 75 to 150 are:
79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149,

Logic To Find Prime Number, using While Loop

In this method, we apply square root to the user entered number and store it inside variable inum. This reduces the number of iterations of inner while loop.

For example,
If num = 100;
inum = sqrt(num);
inum = sqrt(100);
inum = 10;

User entered number 100 is perfectly divisible by 5 and 10, so number 100 is not a prime number.

So its enough if we iterate through the while loop sqrt(num) times to check if the selected number is divisible by any number other than 1 and itself.

Table of all prime numbers up to 1,000:

prime number or not

Note: We are not using curly braces around if statement because we only have 1 line of code after if – so curly braces are optional. If we have multiple lines of code, then we must use curly braces to wrap around the block of code.

For list of all c programming interviews / viva question and answers visit: C Programming Interview / Viva Q&A List

For full C programming language free video tutorial list visit:C Programming: Beginner To Advance To Expert

C Program To Find Prime Numbers From 2 To N, using While Loop

Lets write a C program to find and print / display all the prime numbers from 2 to N. Here N is the user entered number / limit.

Prime Number: Any natural number which is greater than 1 and has only two factors i.e., 1 and the number itself is called a prime number.

Video Tutorial: C Program To Find Prime Numbers From 2 To N, using While Loop

[youtube https://www.youtube.com/watch?v=iYHo-9rClAs]

YouTube Link: https://www.youtube.com/watch?v=iYHo-9rClAs [Watch the Video In Full Screen.]

While loop Logic

All the numbers are perfectly divisible by number 1, so we initialize the variables start and count to 2, instead of 1. So our c program starts checking for divisibility from number 2.

Source Code: C Program To Find Prime Numbers From 2 To N, using Nested While Loop

#include < stdio.h >
#include < math.h >

int main()
{
    int start = 2, num, count, flag, inum;

    printf("Enter the limit\n");
    scanf("%d", &num);

    printf("\n\nPrime numbers from 2 to %d are:\n", num);

    while(start <= num)
    {
        inum = sqrt(start);
        count= 2;
        flag = 1;

        while(count <= inum)
        {
            if(start%count == 0)
            {
                flag = 0;
                break;
            }
            count++;
        }

        if(flag) printf("%d, ", start);

        start++;
    }

    printf("\n\n");

    return 0;
}

Output 1:
Enter the limit
50

Prime numbers from 2 to 50 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,

Output 2:
Enter the limit
75

Prime numbers from 2 to 75 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73,

Output 3:
Enter the limit
100

Prime numbers from 2 to 100 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,

Output 4:
Enter the limit
500

Prime numbers from 2 to 500 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499,

Output 5:
Enter the limit
1000

Prime numbers from 2 to 1000 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997,

Logic To Find Prime Numbers From 2 To N, using While Loop

In this method, we apply square root to the user entered number and store it inside variable inum. This reduces the number of iterations of inner while loop.

For example,
If num = 100;
inum = sqrt(num);
inum = sqrt(100);
inum = 10;

User entered number 100 is perfectly divisible by 5 and 10, so number 100 is not a prime number.

So its enough if we iterate through the while loop sqrt(num) times to check if the user entered number is divisible by any number other than 1 and itself.

We start checking for prime number from number 2 till the user entered number. Variable start is assigned an initial value of 2. For each iteration of outer while loop value of start increments by 1. Inside the inner while loop we check if the value present in variable start is prime number or not. If its prime number then we print that number on to the console window.

Outer while loop selects the number and stores it in variable start on each iteration. Inner while loop checks if the selected number(present in variable start) is prime number or not. You can know the complete logic to check whether a selected number is prime or not: C Program To Find Prime Number or Not using While Loop

Table of all prime numbers up to 1,000:

prime number or not

For list of all c programming interviews / viva question and answers visit: C Programming Interview / Viva Q&A List

For full C programming language free video tutorial list visit:C Programming: Beginner To Advance To Expert