Example 2 Given $0<a<1$, and satisfies
$$
\left[a+\frac{1}{30}\right]+\left[a+\frac{2}{30}\right]+\cdots+\left[a+\frac{29}{30}\right]=18 \text {. }
$$

Then $[10 a]=$ $\qquad$
Solve: 
$$
\begin{array}{l}
0<a<1 \\
\Rightarrow 0<a+\frac{1}{30}<a+\frac{2}{30}<\cdots<a+\frac{29}{30}<2 \\
\Rightarrow\left[a+\frac{1}{30}\right],\left[a+\frac{2}{30}\right], \cdots,\left[a+\frac{29}{30}\right] \text { is either }
\end{array}
$$

0 or 1.
By the problem, we know that 18 of them are equal to 1, and 11 are equal to 0.
$$
\begin{array}{l}
\text { Therefore, }\left[a+\frac{1}{30}\right]=\left[a+\frac{2}{30}\right]=\cdots=\left[a+\frac{11}{30}\right]=0, \\
{\left[a+\frac{12}{30}\right]=\left[a+\frac{13}{30}\right]=\cdots=\left[a+\frac{29}{30}\right]=1 .}
\end{array}
$$

Thus, $0<a+\frac{11}{30}<1,1 \leqslant a+\frac{12}{30}<2$.
Solving this, we get $18 \leqslant 30 a<1 \Rightarrow 6 \leqslant 10 a<\frac{19}{3}$.
Therefore, $[10 a]=6$.