This is not homework.

I saw this article praising Linq library and how great it is for doing combinatorics stuff, and I thought to myself: Python can do it in a more readable fashion.

After half hour of dabbing with Python I failed. Please finish where I left off. Also, do it in the most Pythonic and efficient way possible please.

`from itertools import permutations`

from operator import mul

from functools import reduce

glob_lst = []

def divisible(n): return (sum(j*10^i for i,j in enumerate(reversed(glob_lst))) % n == 0)

oneToNine = list(range(1, 10))

twoToNine = oneToNine[1:]

for perm in permutations(oneToNine, 9):

for n in twoToNine:

glob_lst = perm[1:n]

#print(glob_lst)

if not divisible(n):

continue

else:

# Is invoked if the loop succeeds

# So, we found the number

print(perm)

Thanks!

Here’s a short solution, using itertools.permutations:

`from itertools import permutations`

def is_solution(seq):

return all(int(seq[:i]) % i == 0 for i in range(2, 9))

for p in permutations('123456789'):

seq = ''.join(p)

if is_solution(seq):

print(seq)

I’ve deliberately omitted the divisibility checks by 1 and by 9, since they’ll always be satisfied.