# Euler Problem 1¶

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

In [1]:

    def get_sum_of_natural_dividable_by_3_and_5_below(m):
        return sum([x for x in range(m) if x % 3 == 0 or x % 5 == 0])

Test example provided in problem statement:

In [2]:

    assert(get_sum_of_natural_dividable_by_3_and_5_below(10) == 23)

In [3]:

    print(get_sum_of_natural_dividable_by_3_and_5_below(1000))

    233168