If inc/exclusion is unknown then, note $\,3,5\mid n\!+\color{blue}{\!15k}\iff 3,5\mid n,\,$ so the multiples of $\,3,5\,$ have periodicity $15,\,$ so we can split the sum into chunks from each period as follows
$$\begin{eqnarray}
\color{blue}{0}+\overbrace{\{0,3,5,6,9,10,12\}}^{\large \rm sum\, =\, \color{#c00}{45}}\\
\color{blue}{15}+\{0,3,5,6,9,10,12\}\\
\color{blue}{30}+\{0,3,5,6,9,10,12\}\\
\cdots\qquad\qquad\\
\color{blue}{15\cdot 46}+\{0,3,5,6,9,10,12\}\\
\underbrace{15\cdot 47}_{\large\color{#0a0}{705}} + \underbrace{15\cdot 47+3}_{\large\color{#0a0}{708}}\qquad\quad\ \ \, \\
\hline
\end{eqnarray}\qquad\qquad$$
with total sum $\, =\, \color{blue}{7\cdot 15\, (47\cdot 46/2)} + 47\cdot \color{#c00}{45}\color{#0a0}{ + 705 + 708} = 117033$