Okay so the problem itself is really simple,
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
Many people will just solve this using a for loop and and the modulo operator like this,
s = 0
for x in range(1000):
if x % 3 == 0 or x % 5 == 0:
s += x
print(s)
But I felt the need to solve it using just mathemathics and came up with this: Let $s$ be the sum, $n=10$ and $x_1 = 3, x_2=5$. We have,
$$ \newcommand\c[1]{\left\lceil#1\right\rceil} \newcommand\p[1]{\left(#1\right)} \newcommand\ep{\varepsilon} $$
\begin{align*} s &= (3 + 6 + 9) + (5) \\ &= x_1(1 + 2 + 3) + x_2(1) \\ &= x_1\sum_{i=1}^{m_1}i + x_2\sum_{i=1}^{m_2}i && m_i = \c{\frac{n}{x_i}}-1 \\ &= x_1\frac{m_1(m_1+1)}{2} + x_2\frac{m_2(m_2+1)}{2} \\ &= x_1\frac{\p{\c{\frac{n}{x_1}}-1}\c{\frac{n}{x_1}}}{2} + x_2\frac{\p{\c{\frac{n}{x_2}}-1}\c{\frac{n}{x_2}}}{2} \\ &= \sum_{i}\p{x_i\frac{\p{\c{\frac{n}{x_i}}-1}\c{\frac{n}{x_i}}}{2}} \end{align*}
At this point, we have over summed since $x_1, x_2$ may share multiples. So if they are relatively prime, we can write:
$$ s = \sum_{i}\p{x_i\frac{\p{\c{\frac{n}{x_i}}-1}\c{\frac{n}{x_i}}}{2}} - \ep $$
where $\ep = x_3\frac{\p{\c{\frac{n}{x_3}}-1}\c{\frac{n}{x_3}}}{2}$ for $x_3 = x_1x_2$. But now the issue is the assumption that $x_i$'s are relatively prime or there is only two of them. Consider a more general case where $x_i \in X \supset \mathbb{N}$ and $x_i < n \in \mathbb{N}$. Is there an elegant way to state $\ep$ in this case? I feel like I need to use LCM somehow but can't quite think how.
Example. Let $X = \{2,3,9\}$ and $n = 10$. We have:
\begin{align*} 2&:\{2,4,6,8\} \\ 3&:\{3,6,9\} \\ 9&:\{9\} \end{align*}
Now, we need a way to count (or have the same effect thereof) the 6 and 9 only once.