#!/usr/bin/env python3

# 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.

from timeit import default_timer


def main():
    start = default_timer()

    sum_ = 0

#   Simple brute-force approach: try every number between 3 and 999,
#   check if it's a multiple of 3 or 5, if yes add it to the total.
    for i in range(3, 1000):
        if i % 3 == 0 or i % 5 == 0:
            sum_ += i

    end = default_timer()

    print('Project Euler, Problem 1')
    print(f'Answer: {sum_}')

    print(f'Elapsed time: {end - start:.9f} seconds')


if __name__ == '__main__':
    main()