1. Sum of naturals pisible by 3 and 5

The solution to this problem is to iterate through all numbers from 3 (1 and 2 are not pisible by 3 so it does not make sense to test them) up to the limit entered by the user. Use the modulo operation to check that the rest of the pision of a number by 3 and 5 is 0. However, the trick to being able to sum up to a larger limit is to use long long and not int or long for the sum, which would result in an overflow before summing up to 100,000:

int main()
{
   unsigned int limit = 0;
   std::cout << "Upper limit:";
   std::cin >> limit;

   unsigned long long sum = 0;
   for (unsigned int i = 3; i < limit; ++i)
   {
     if (i % 3 == 0 || i % 5 == 0)
        sum += i;
   }

   std::cout << "sum=" << sum << std::endl;
}