My code in Julia, almost identical as the Python code (see below), runs in 4.6 s while the Python version runs in 2.4 s. Obviously there is a lot or room for improvement.

```
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 below:

```
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)
```