Question

8.(a) Write a program that prints all of the numbers from 0 to 102 divisible by either 3 or 7.

​ (b) Write a program that prints all of the numbers from 0 to 102 divisible by both 3 and 7

(c) Write a program that prints all of the even numbers from 0 to 102 divisible by both 3 and 7

(d) Write a program that prints all of the odd numbers from 0 to 102 divisible by both 3 and 7

(e) Write a program that prints all of the even numbers from 30 to 102 divisible by either 3 or 7

(f) Write a program that prints all of the odd numbers from 28 to 102 divisible by either 3 or 7

(g) Write a program that prints all of the odd numbers between 30 to 102 divisible by both 3

​​and 7. in loops

Answer 1

(a) Write a program that prints all of the numbers from 0 to 102 divisible by either 3 or 7.
Program:

#include <stdio.h>

int main()
{
int i = 0;
       printf("Numbers from 0 to 102 divisible by either 3 or 7 : ");
       for (i = 0; i <= 102; i++) {
           if ((i % 3 == 0) || (i % 7 == 0)) {
               printf("%d\t",i);
           }
       }
      

return 0;
}

Output:
Numbers from 0 to 102 divisible by either 3 or 7 : 0   3   6   7   9   12   14   15   18   21   24   27   28   30   33   35   36   39   42   45   48   49   51   54   56   57   60   63   66   69   70   72   75   77   78   81   84   87   90   91   93   96   98   99   102  

(b) Write a program that prints all of the numbers from 0 to 102 divisible by both 3 and 7
Program:

#include <stdio.h>

int main()
{
int i = 0;
       printf("\nNumbers from 0 to 102 divisible by either 3 and 7 : ");
       for (i = 0; i <= 102; i++) {
           if ((i % 3 == 0) && (i % 7 == 0)) {
               printf("%d\t",i);
           }
       }
      
return 0;
}
Output:
Numbers from 0 to 102 divisible by either 3 and 7 : 0   21   42   63   84  

(c) Write a program that prints all of the even numbers from 0 to 102 divisible by both 3 and 7
Program:

#include <stdio.h>
int main()
{
int i = 0;
       printf("\nEven numbers from 0 to 102 divisible by either 3 and 7 : ");
       for (i = 0; i <= 102; i++) {
           if (((i % 3 == 0) && (i % 7 == 0)) && (i % 2 == 0)) {
               printf("%d\t",i);
           }
       }  

return 0;
}
Output:
Even numbers from 0 to 102 divisible by either 3 and 7 : 0   42   84  
(d) Write a program that prints all of the odd numbers from 0 to 102 divisible by both 3 and 7
Program:

#include <stdio.h>
int main()
{
int i = 0;
       printf("\nOdd numbers from 0 to 102 divisible by either 3 and 7 : ");
       for (i = 0; i <= 102; i++) {
           if (((i % 3 == 0) && (i % 7 == 0)) && (i % 2 != 0)) {
               printf("%d\t",i);
           }
       }  

return 0;
}
Output:
Odd numbers from 0 to 102 divisible by either 3 and 7 : 21   63  

(e) Write a program that prints all of the even numbers from 30 to 102 divisible by either 3 or 7
Program:

#include <stdio.h>
int main()
{
int i = 0;
       printf("\nEven numbers from 30 to 102 divisible by either 3 or 7 : ");
       for (i = 30; i <= 102; i++) {
           if (((i % 3 == 0) || (i % 7 == 0)) && (i % 2 == 0)) {
               printf("%d\t",i);
           }
       }  

return 0;
}
Output:
Even numbers from 30 to 102 divisible by either 3 or 7 : 30   36   42   48   54   56   60   66   70   72   78   84   90   96   98   102  

(f) Write a program that prints all of the odd numbers from 28 to 102 divisible by either 3 or 7
Program:

#include <stdio.h>
int main()
{
int i = 0;
       printf("\nOdd numbers from 28 to 102 divisible by either 3 or 7 : ");
       for (i = 28; i <= 102; i++) {
           if (((i % 3 == 0) || (i % 7 == 0)) && (i % 2 != 0)) {
               printf("%d\t",i);
           }
       }

return 0;
}
Output:
Odd numbers from 28 to 102 divisible by either 3 or 7 : 33   35   39   45   49   51   57   63   69   75   77   81   87   91   93   99  

(g) Write a program that prints all of the odd numbers between 30 to 102 divisible by both 3
Program:

#include <stdio.h>
int main()
{
int i = 0;
       printf("\nOdd numbers from 30 to 102 divisible by either 3 and 7 : ");
       for (i = 30; i <= 102; i++) {
           if (((i % 3 == 0) && (i % 7 == 0)) && (i % 2 != 0)) {
               printf("%d\t",i);
           }
       }
return 0;
}
Output:
Odd numbers from 30 to 102 divisible by either 3 and 7 : 63  

Explanation:

Here I have used the "% (modulus)" operator to check whether number is divisible by 3 or 7 and if gives us the remainder after dividing
Example:
21%3 --> 0 it means 21 is divisible by 3

same % operator is used depending on the required result in each program.