I had read that the modulo operator (remainder after integer division) was slow but had never had a chance to try to break it myself, until a couple of days ago. I was distracted from my current project by Project Euler, after searching for a way to test and improve my programming and maths skills.

Project Euler is a collection of problems designed to exercise those skills and its first, simplest task is to add up the multiples of 3 and 5 below 1,000, which I first solved with a fairly typical “brute-force” approach:

sum = 0
for x in range(1, 1000):
    if x % 3 == 0 or x % 5 == 0:
        sum += x

This iterates over the numbers 1 through 999, checking to see if there’s a remainder after dividing by 3 or 5. After finishing, I read on the problem’s forum thread of a way of achieving the result with a little more maths and without using the modulo.

I tried it out for fun and to be honest wasn’t impressed — when timed with the Unix time command both methods took around 0.025 seconds.

So I decided to kick the limit up a bit, from 999 to 999,999,999.

The modulo method took 4 minutes and 24.506 seconds (264.506s) while the more mathematical approach held steady at 0.026 seconds. Over 10,000 times as fast. Yes, I double-checked the results.

The modulo code was similar to that above while the other method was written like so:

def sum_series(multiple, top_limit):
    mult_max = top_limit // multiple
    return multiple * (mult_max * (mult_max + 1)) / 2

print(sum_series(3, 999)
        + sum_series(5, 999)
        - sum_series(15, 999))

The sum_series function works out how many multiples of a given number there are between 0 and top_limit and adds them up.

But you can avoid iterating over them because of the interesting fact that the sum of all natural numbers through x is x × (x + 1) ÷ 2. It took me a while of head-scratching before I read an explanation (two, actually) that made sense.

The function does that for the number of multiples, and then multiplies the result by the number in question to actually make them multiples. For example, 1 + 2 + 3 through 333 become 3 + 6 + 9 through 999.

Repeat with 5, then subtract the results for 15 (to remove the double-counted multiples of both 3 and 5) and you’re done.