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: introductory

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==0for x,y in __import__('itertools').permutations(s.split(),2))

from itertools import permutations
def solve(s,k):
    return sum(int(a + b) % k == 0 for a, b in 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