66. The sum

$$
\begin{gathered}
2 \\
3 \cdot 6
\end{gathered}+\begin{gathered}
2 \cdot 5 \\
3 \cdot 6 \cdot 9
\end{gathered}+\ldots+\begin{gathered}
2 \cdot 5 \cdot \ldots \cdot 2015 \\
3 \cdot 6 \cdot \ldots \cdot 2019
\end{gathered}
$$

was written as a decimal fraction. Find the first digit after the decimal point.
66. Answer: the first digit after the decimal point is 5.

To start, let's simplify the given sum. Each term can be written as a difference

$$
\begin{aligned}
\frac{2 \cdot 5 \cdot \ldots \cdot(3 k-1)}{3 \cdot 6 \cdot 9 \cdot \ldots \cdot(3 k+3)}=\frac{2 \cdot 5 \cdot \ldots \cdot(3 k-1) \cdot(3 k+3)}{3 \cdot 6 \cdot 9 \cdot \ldots \cdot(3 k+3)}- & \frac{2 \cdot 5 \cdot \ldots \cdot(3 k-1) \cdot(3 k+2)}{3 \cdot 6 \cdot 9 \cdot \ldots \cdot(3 k+3)}= \\
& =\frac{2 \cdot 5 \cdot \ldots \cdot(3 k-1)}{3 \cdot 6 \cdot 9 \cdot \ldots \cdot 3 k}-\frac{2 \cdot 5 \cdot \ldots \cdot(3 k-1) \cdot(3 k+2)}{3 \cdot 6 \cdot 9 \cdot \ldots \cdot(3 k+3)}
\end{aligned}
$$

Then the entire sum will telescope to the difference of the extreme terms $\frac{2}{3}-\frac{2 \cdot 5 \cdot \ldots \cdot 2018}{3 \cdot 6 \cdot \ldots \cdot 2019}$.

Now let's estimate the subtrahend. Introduce the variables

$$
\begin{gathered}
A=\frac{1 \cdot 3 \cdot 6 \cdot \ldots \cdot 2016}{1 \cdot 4 \cdot 7 \cdot \ldots \cdot 2017}, \quad B=\frac{1 \cdot 4 \cdot 7 \cdot \ldots \cdot 2017}{2 \cdot 5 \cdot 8 \cdot \ldots \cdot 2018}, \quad C=\frac{2 \cdot 5 \cdot 8 \cdot \ldots \cdot 2018}{3 \cdot 6 \cdot 9 \cdot \ldots \cdot 2019}, \quad D=\frac{3 \cdot 6 \cdot 9 \cdot \ldots \cdot 2019}{4 \cdot 7 \cdot 10 \cdot \ldots \cdot 2020} \\
E=\frac{4 \cdot 7 \cdot 10 \cdot \ldots \cdot 2020}{5 \cdot 8 \cdot 11 \cdot \ldots \cdot 2022}
\end{gathered}
$$

We want to estimate the value of the number $C$.

Since $\frac{a-1}{a}<\frac{a}{a+1}$ for natural $a$, the inequalities $A<B<C<D<E$ hold, from which $A B C<C^{3}<C D E$. Substituting these inequalities into the formulas for our numbers and simplifying the fractions, we get

$$
\frac{1}{2019}<C^{3}<\frac{2}{2022}, \quad \frac{1}{15}<\sqrt[3]{\frac{1}{2019}}<C<\sqrt[3]{\frac{2}{2022}}<\frac{1}{6}, \text { and thus, } \quad \frac{1}{2}<\frac{2}{3}-\frac{2 \cdot 5 \cdot \ldots \cdot 2018}{3 \cdot 6 \cdot \ldots \cdot 2019}<\frac{3}{5}
$$

Thus, the first digit after the decimal point of the original number is 5.