We have two consecutive integers k1 and k2, k2 = k1 + 1

We need to calculate the lowest integer `n`, such that:
the values nk1 and nk2 have the same digits but in different order.

E.g.# 1:
```
k1 = 100
k2 = 101
n = 8919
#Because 8919 * 100 = 891900 
and      8919 * 101 = 900819
```

E.g.# 2:
```
k1 = 325
k2 = 326
n = 477
#Because 477 * 325 = 155025
and      477 * 326 = 155502
```

Your task is to prepare a function that will receive the value of `k` and outputs the value of `n`.

The examples given above will be:
```python
find_lowest_int(100) === 8919
find_lowest_int(325) ===  477
```
Features of the random tests
```
10 < k < 10.000.000.000.000.000 (For Python, Ruby and Haskell)
10 < k < 1.000.000.000  (For Javascript 1e9)
```
Enjoy it!!

Ruby and Javascript versions will be released soon.

Difficulty: introductory

Starter code:

def find_lowest_int(k):
	

Solutions:
def find_lowest_int(k1):
    k2, n = k1 + 1, 1

    def digits(n):
        return sorted(str(n))
    
    while digits(n*k1) != digits(n*k2):
        n += 1
    
    return n

from itertools import count as c

def find_lowest_int(k):
    return next((n for n in c(1) if sorted(str(n*k))== sorted(str(n*(k+1)))))



def find_lowest_int(k):
    return next(n for n in range(9, 9999999, 9) if sorted(str(n * k)) == sorted(str(n * (k+1))))

def find_lowest_int(k):
    l, n = k + 1, 9
    while digits(k * n) != digits(l * n):
        n += 9
    return n

def digits(n):
    return sorted(str(n))    

def find_lowest_int(number):
    multiplier = 1
    while sorted(str(number * multiplier)) != sorted(str((number + 1) * multiplier)):
        multiplier += 1
    return multiplier

from itertools import count

def find_lowest_int(k):
    return next(n for n in count(1) if sorted(str(n*k)) == sorted(str(n*(k+1))))

from collections import Counter

# idea: the difference between k*n and (k+1)*n is n
# for them to have the same digits they must have the same digit sum
# so n must have a digit sum of 0 (mod 9) - n must be divisible by 9
def find_lowest_int(k):
    n = 9
    while True:
        if Counter(str(k*n)) == Counter(str((k+1)*n)):
            return n
        n += 9


def find_lowest_int(k):
    n=2
    while True:
        if sorted(str(n*k))==sorted(str(n*(k+1))):return n
        n+=1