Thanks bro!

• two loops, but not imbricated : sum
• counting the number of iterations of each loop: n/3+n/5 = 8n/15 "~=" n/2
• asymptotic behavior: O(n)

EDIT: or you just go the raw way... x)

• one loop ~= O(n)
• 2 consecutive loops = O(n)+O(n) = 2*O(n) => O(n)

Interesting. How would you calculate that?

nope: O(n/3)+O(n/5) so around O(n/2), so finally O(n)

That is an o(n**2) solution correct?

Yeah I should've come up with some better name but you got the point.
No harsh feeling :)

Lies. sum5 isn't sum of multiples of 5 but sum of multiples of 5 minus multiples of 15.

This is so concise and clever!

Clever solution!