Consider the string `"1 2 36 4 8"`. Lets take pairs of these numbers, concatenate each pair and determine how many of them of divisible by `k`. 

```Pearl 
If k = 3, we get following numbers ['12', '18', '21', '24', '42', '48', '81', '84'], all divisible by 3.   
Note:
-- 21 and 12 are different pairs. 
-- Elements must be from different indices, so '3636` is not a valid. 
```

Given a string of numbers and an integer `k`, return the number of pairs that when concatenated, are divisible by `k`.

```
solve("1 2 36 4 8", 3) = 8, because they are ['12', '18', '21', '24', '42', '48', '81', '84']
solve("1 3 6 3", 3) = 6. They are ['36', '33', '63', '63', '33', '36']
```

More examples in test cases. Good luck!

Please also try [Simple remove duplicates](https://www.codewars.com/kata/5ba38ba180824a86850000f7)

Difficulty: EASY
Starter code: def solve(s,k):
	

Solutions:
from itertools import permutations

def solve(s, k):
	return sum((not v % k for v in map(int, map(''.join, permutations(s.split(), 2)))))


from itertools import permutations

def solve(stg, k):
	return sum((1 for (m, n) in permutations(stg.split(), 2) if int(f'{m}{n}') % k == 0))


from itertools import permutations

def solve(s, k):
	return sum((not int(a + b) % k for (a, b) in permutations(s.split(), 2)))


from itertools import permutations as p

def solve(s, k):
	return len([j for j in [int(''.join(i)) for i in p(s.split(), 2)] if j % k == 0])


solve = lambda s, k: sum((int(x + y) % k == 0 for (x, y) in __import__('itertools').permutations(s.split(), 2)))


from itertools import permutations, filterfalse

def solve(s, k):
	return len(list(filterfalse(lambda x: int(''.join(x)) % k, permutations(s.split(), 2))))


import itertools as it

def solve(s, k):
	return sum((n % k == 0 for n in map(int, map(''.join, it.permutations(s.split(), 2)))))


def solve(s, k):
	s = s.split()
	return sum((1 for i in range(len(s)) for j in range(len(s)) if i != j and int(s[i] + s[j]) % k == 0))


def solve(s, k):
	count = 0
	l = s.split()
	for i in range(len(l) - 1):
		for j in range(i + 1, len(l)):
			count += int(int(l[i] + l[j]) % k == 0) + int(int(l[j] + l[i]) % k == 0)
	return count