The solution to this problem is to iterate through all numbers from 3 (1 and 2 are not divisible 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 division 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;}
