Problem

Given a list of numbers, return the sum of all multiples of n or m

i.e. n = 3, m = 5, numbers = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]

Listing all the multiples of 3 and 5 gives us [ 3, 5, 6, 9 ] with a sum of 23

We can easily check if a number, x, is a multiple of y if x can divide y without any remainders. For this, we can use the modulo operator.

For example:

# test if a number is divisible by 3
x % 3 == 0

# test if a number is divisible by 5
x % 5 == 0

Now, with this in mind we can get all multiples of 3 and 5. Using on our example:

numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9]
[e for e in numbers if e % 5 == 0 or e % 3 == 0]

[3, 5, 6, 9]

This gives us the expected [3, 5, 6, 9] list of numbers. Putting it all together:

from typing import List


def sum_of_multiples(n: int, m: int, numbers: List):
    multiples = [e for e in numbers if e % n == 0 or e % m == 0]
    return sum(multiples)

assert sum_of_multiples(3, 5, [1, 2, 3, 4, 5, 6, 7, 8, 9]) == 23

Problem (variant)

Given an upper bound, generate an array of positive integers less than the upper bound and return the sum of all multiples of n or m

def sum_of_multiples2(n: int, m: int, upper_bound: int):
    multiples = [e for e in range(upper_bound) if e % n == 0 or e % m == 0]
    return sum(multiples)

assert sum_of_multiples2(3, 5, 1000) == 233168