$x \leq 10000$

Basically we have to sum all the multiples of $3$ less than $10000$ with all the multiples of $11$ less than $10000$ and subtract $2$ times all the multiples of $33$ less than $10000$ (which were added two times: one in the summation of the multiples of $3$ and one in the summation of the multiples of $11$ )

We first have to count how many multiples of $3$ , $11$ and $33$ there are less than $10000$ , which are respectively

• $\left\lfloor\dfrac{10000}{3}\right\rfloor=3333$

• $\left\lfloor\dfrac{10000}{11}\right\rfloor=909$

• $\left\lfloor\dfrac{10000}{33}\right\rfloor=303$

Hence the required sum is

$\begin{aligned} S&=\displaystyle\sum_{i=1}^{3333}3i+\displaystyle\sum_{j=1}^{909}11j-2\displaystyle\sum_{k=1}^{303}33k\\ &=3\displaystyle\sum_{i=1}^{3333}i+11\displaystyle\sum_{j=1}^{909}j-66\displaystyle\sum_{k=1}^{303}k\\ &=3\cdot\dfrac{3333\cdot 3334}{2}+11\cdot\dfrac{909\cdot 910}{2}-66\cdot\dfrac{303\cdot 304}{2}\\ &=\boxed{18178182} \end{aligned}$