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.

Source: codewars | Difficulty: EASY
Tags: Fundamentals

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