Your task it to return ```true``` if the fractional part (rounded to 1 digit) of the result (```a``` / ```b```) exists, more than ```0``` and is multiple of ```n```.
Otherwise return ```false```. (For Python return True or False)

All arguments are positive digital numbers.

Rounding works like toFixed() method. (if more than...5 rounds up)

Find exapmles below: 

```

isMultiple(5, 2, 3) -> false // 2.5 -> 5 is not multiple of 3
isMultiple(5, 3, 4) -> false // 1.7 -> 7 is not multiple of 4
isMultiple(5, 4, 3) -> true // 1.3 -> 3 is multiple of 3
isMultiple(666, 665, 2) -> false // 1.0 -> return false


```

Source: codewars | Difficulty: EASY
Tags: Fundamentals

Starter code: def isMultiple(a, b, n):
	

Solutions:
def isMultiple(a, b, n):
	remainder = int((a / b + 0.05) * 10) % 10
	return remainder > 0 and remainder % n == 0


def isMultiple(a, b, n):
	frac_part = int(10.0 * (a / b - a // b) + 0.5) % 10
	return frac_part != 0 and frac_part % n == 0


isMultiple = lambda a, b, n: (lambda d: d > 0 == d % n)((20 * a / b + 1) // 2 % 10)


def isMultiple(a, b, n):
	x = round(float('.' + str(a / b + 1e-05).split('.')[-1]), 1)
	x = x * 10 if x < 1 else x
	return x != 0 and x % n == 0


def isMultiple(a, b, n):
	c = round(a / b + 0.0001, 1) * 10 % 10
	return c % n == 0 and bool(c)


from math import floor

def isMultiple(a, b, n):
	fractional_part = floor(10 * (a / b - a // b) + 0.5)
	if fractional_part == 10:
		fractional_part //= 10
	return fractional_part and fractional_part % n == 0


def isMultiple(a, b, n):
	x = int((a / b - a // b) * 10 + 0.5)
	return 0 < x < 10 and x % n == 0


import math

def normal_round(n):
	if n - math.floor(n) < 0.5:
		return math.floor(n)
	return math.ceil(n)

def isMultiple(a, b, n):
	to_test = normal_round((a / b - int(a / b)) * 10)
	if to_test >= 10:
		return False
	else:
		return to_test > 0 and to_test % n == 0


def isMultiple(a, b, n):

	def rond(x):
		return int(x) if x % 1 < 0.5 else int(x + 1)
	res = a / b
	res = rond(res * 10) / 10
	remainder = round(res % 1, 1)
	return remainder and (not 10 * remainder % n)


def isMultiple(a, b, n):
	print(a / b)
	print(n)
	quot = a / b * 10
	digit = int(quot) % 10
	if quot % 1 >= 0.5:
		digit = (int(quot) + 1) % 10
	return digit > 0 and digit % n == 0