Returning the sum of all multiples of a number up to 1000
range takes an optional third argument, which is the step size for the sequence. For example:

range(0, 10, 3) # Returns [0, 3, 6, 9]


Python also has a built-in sum function that adds up all the numbers in an array, so you could refactor your code to look like this:

def get_multiple_sum(num):
    return sum(range(0, 1000, num))


Performance-wise, this should be much faster, as it doesn't perform 1000 remainder operations and uses a library function which may be optimized by the interpreter.