Brute force

The first problem is actually quite easy, the naive solution is to iterate over each number between 0 and 1000 and check those that are multiples of 3 or 5.

From solution1.py:

def sum_of_three_and_five(limit=1000):
    return sum(i for i in range(limit) if i % 3 == 0 or i % 5 == 0)