Comparing Julia and Python for basic operations

General Usage Performance
3

Thx for your answer!
I’ve made a couple of changes (I’m not interested in the divisors themselves, just in the number of them) and now the code runs in 0.12 s :smiley:

    function num_divisors(n)
        #res = floor(sqrt(n))
        res = floor(Int, sqrt(n))
        divs = 0#[]
        for i = 1:res
            if n % i == 0
                divs += 1
                #append!(divs, i)
            end
        end
        if res^2 == n
            divs -= 1
            #pop!(divs)
        end
        return 2 * divs #length(divs)
    end

Regarding the validity of the function, I guess that it’s correct as it gives the solution as good in Project Euler.