Question

#include <iostream>
using namespace std;

int main() {
   int sumOdd = 0; // For accumulating odd numbers, init to 0
   int sumEven = 0; // For accumulating even numbers, init to 0
   int upperbound; // Sum from 1 to this upperbound

   // Prompt user for an upperbound
   cout << "Enter the upperbound: ";
   cin >> upperbound;

   // Use a loop to repeatedly add 1, 2, 3,..., up to upperbound
   int number = 1;
   while (number <= upperbound) {
      if (number % 2 == 0) { // even number
         sumEven = sumEven + number;
      } else {                // odd number
         sumOdd = sumOdd + number;
      }
      ++number; // increment number by 1
   }

   // Print the results
   cout << "The sum of odd numbers is " << sumOdd << endl;
   cout << "The sum of even numbers is " << sumEven << endl;
   cout << "The difference is " << (sumOdd - sumEven) << endl;

   return 0;
}

QUESTION:

-Write a program to sum all the integers between 1 and 1000, that are divisible by 13, 15 or 17, but not by 30.

Answer 1

#include <iostream>
using namespace std;

int main() {
int sum = 0; // For accumulating odd numbers, init to 0
  

// Use a loop to repeatedly add 1, 2, 3,..., up to 1000
for(int i=1;i<=1000;i++){
// checking if it is divisible by 13 and 15 ot by 17 and not by 30
if( ((i % 13==0 && i%15==0 ) || (i%17==0)) && i%30!=0)
sum+=i;
}
  
// Print the results
cout << "The sum of odd numbers that are divisible by 13, 15 or 17, but not by 30. " << sum << endl;
return 0;
}