Solving Project Euler Problem 1 using extension methods
This is not so much of a code answer as a mathematical answer.

We can calculate directly the number of multiples \$m\$ of a natural number \$n\$ that are below the limit \$x\$.  It's simply \$\lceil x/n\rceil\$.  The sum of those multiples is:
$$n\sum_{i=1}^{\lceil x/n\rceil}= n\left(\frac{1}{2}(\lceil x/n \rceil)(\lceil x/n\rceil+1)\right)$$
So for \$n=3\$ and \$x=999\$, we get $$3\left(\frac{1}{2}(333)(333+1)\right)= 166833$$
And for \$n=5\$ and \$x=999\$, we get $$5\left(\frac{1}{2}(200)(200+1)\right)=99500$$
So we could calculate the answer is $$166833 + 99500 = 266333$$  However the problem here (hat tip to @MartinR !) is that we have double-counted multiples of 15.  To fix that, we need to subtract that count:

$$ 15\left(\frac{1}{2}(66)(66+1)\right) = 33165$$
So the correct final answer is $$266333 - 33165 = 233168$$