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


```

Difficulty: introductory

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+.00001).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
    #Extra round here to deal with floating-point nonsense
    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