3. Let $\lfloor x\rfloor$ denote the greatest integer not exceeding the real number $x$. Find the integer solution to the equation
$$
\left\lfloor\frac{x}{1!}\right\rfloor+\left\lfloor\frac{x}{2!}\right\rfloor+\cdots+\left\lfloor\frac{x}{10!}\right\rfloor=2019
$$
3. Clearly, $x \leqslant 2018$.

Without loss of generality, move the last four terms on the left side of the original equation to the right side. Assume
$$
\begin{aligned}
x= & a(6!)+b(5!)+c(4!)+ \\
& d(3!)+e(2!)+f(1!),
\end{aligned}
$$

where $a, b, c, d, e, f \in \mathbf{Z}_{+}$, and $a \leqslant 6, b \leqslant 5$, $c \leqslant 4, d \leqslant 3, e \leqslant 2, f \leqslant 1$.
Substitute equation (1) into the original equation to get
$$
\begin{array}{l}
1237 a+206 b+41 c+10 d+3 e+f \\
=2019 .
\end{array}
$$
$$
\begin{array}{l}
\text { By } 206 b+41 c+10 d+3 e+f \\
\leqslant 206 \times 5+41 \times 4+10 \times 3+3 \times 2+1 \\
=1030+164+30+6+1 \\
=1231 \\
\Rightarrow 788 \leqslant 1237 a \leqslant 2019 \\
\Rightarrow a=1 \\
\Rightarrow 206 b+41 c+10 d+3 e+f=782 .
\end{array}
$$

Similarly, we solve $b=3, c=4, d=e=f=0$.
Thus, $x=1(6!)+3(5!)+4(4!)=1176$.
Verification shows
$$
\begin{array}{l}
\left\lfloor\frac{1176}{1!}\right\rfloor+\left\lfloor\frac{1176}{2!}\right\rfloor+\left\lfloor\frac{1176}{3!}\right\rfloor+ \\
\left\lfloor\frac{1176}{4!}\right\rfloor+\left\lfloor\frac{1176}{5!}\right\rfloor+\left\lfloor\frac{1176}{6!}\right\rfloor \\
=1176+588+196+49+9+1 \\
=2019 .
\end{array}
$$