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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 foo(n):
    total = 0
    for i in range(n):
        if i % 3 == 0 or i % 5 == 0:
            total += i
    return total



In [2]:

    
n = 1000
%timeit foo(n)
foo(n)









    



1000 loops, best of 3: 282 us per loop






    Out[2]:





233168



In [3]:

    
def foo(n):
    total = 0
    for i in range(n):
        if i % 3 == 0:
            total += i
        if i % 5 == 0:
            total += i
    return total



In [4]:

    
n = 1000
%timeit foo(n)
foo(n)









    



1000 loops, best of 3: 319 us per loop






    Out[4]:





266333



In [5]:

    
def foo(n):
    return sum(filter((lambda x: x % 3 == 0 or x % 5 == 0), range(n)))



In [6]:

    
n = 1000
%timeit foo(n)
foo(n)









    



1000 loops, best of 3: 384 us per loop






    Out[6]:





233168