#  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.


def f_timer(passed_f):

	def inner_f(*args, **kwargs):

		import time
		t1 = time.time()
		value = passed_f(*args, **kwargs)
		t2 = time.time()
		t3 = t2 - t1
		print("The function {} ran for {:4f} seconds... ".format(passed_f.__name__,t3))
		return value
	
	return inner_f

def check_divisibility(num, dict_multiples: dict) -> bool:
	
	dummy = 1
	
	for m in dict_multiples:
		dummy *= num%m	
	if dummy == 0:
		return True
	else:
		return False

@f_timer
def main():

	multiples = {3,5}
	sumval = 0

	for n in range(1000):
		r = check_divisibility(n,multiples)
		if r:
			sumval += n

	print(sumval)

main()