The number `198` has the property that `198 = 11 + 99 + 88, i.e., if each of its digits is concatenated twice and then summed, the result will be the original number`. It turns out that `198` is the only number with this property. However, the property can be generalized so that each digit is concatenated `n` times and then summed. 

eg:-
```
original number =2997 , n=3
2997 = 222+999+999+777 and here each digit is concatenated three times.

original number=-2997 , n=3

-2997 = -222-999-999-777 and here each digit is concatenated three times.


```
Write afunction named `check_concatenated_sum` that tests if a number has this generalized property.

Difficulty: introductory

Starter code:

def check_concatenated_sum(num, n):
	

Solutions:
def check_concatenated_sum(n, r):
    return abs(n) == sum(int(e*r) for e in str(abs(n)) if r) 

def check_concatenated_sum(number, n):
    return (10**n - 1)//9 * sum(map(int, str(abs(number)))) == abs(number)

def check_concatenated_sum(num, n):
    return n and num % int('1' * n) == 0

def check_concatenated_sum(num, n):
    m = abs(num)
    return m == sum(map(int, str(m))) * int('1' * n or 0)

def check_concatenated_sum(num, l):
    return sum([int(x*l) for x in str(abs(num))]) == abs(num) if l>0 else False

def check_concatenated_sum(number, k):
    return sum(int(d * k) for d in str(abs(number))) == abs(number) if k else False


def check_concatenated_sum(num, n):
    num = abs(num)
    return sum(int(i*n) for i in str(num) if n)==num

def check_concatenated_sum(a,b):
    if not b:return False
    c=abs(a)
    d=[int(f"{(a<0)*'-'}{n*b}") for n in str(c)]
    return sum(d)==a