#  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.

import functools
import time

def timer(f):
	""" Decorator function to calculate compute time of passed function. """
	
	@functools.wraps(f)	
	def wrapper_timer(*args,**kwargs):
		
		start_time = time.perf_counter()

    	f(*args, **kwargs)
    	
    	end_time = time.perf_counter()
    	run_time = end_time - start_time
    	
    	print("Finished {func.__name__!r} in {run_time:.4f} seconds...")

    	return f

	return wrapper_timer

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

@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()