Hi there Julia community!,

I’m quite new to the language and I’m falling in love more and more with it every day

In order to improve my knowledge, I’m revisiting Project Euler problems, which I’ve done in Python, with Julia and, so far, Julia code is waaay faster than Python (who knew! :D)

But I’m getting problems with the following problem that runs in 4.6 s while the Python version runs in 2.4 s. Both versions are almost identical

Julia version

```
function Problem12()
#=
The sequence of triangle numbers is generated by adding the natural
numbers. So the 7th triangle number would be:
1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred
divisors?
=#
function num_divisors(n)
res = floor(sqrt(n))
divs = []
for i in 1:res
if n%i == 0
append!(divs,i)
end
end
if res^2 == n
pop!(divs)
end
return 2*length(divs)
end
triangle = 0
for i in Iterators.countfrom(1)
triangle += i
if num_divisors(triangle) > 500
return string(triangle)
end
end
end
```

Python version (for comparison)

```
import itertools
from math import sqrt, floor
# Returns the number of integers in the range [1, n] that divide n.
def num_divisors(n):
end = floor(sqrt(n))
divs = []
for i in range(1, end + 1):
if n % i == 0:
divs.append(i)
if end**2 == n:
divs.pop()
return 2*len(divs)
def compute():
triangle = 0
for i in itertools.count(1):
# This is the ith triangle number, i.e. num = 1 + 2 + ... + i =
# = i*(i+1)/2
triangle += i
if num_divisors(triangle) > 500:
return str(triangle)
```