451986 can be written in the form: $1986=a_{1}+a_{2} \cdot 2!+a_{3} \cdot 3!+\cdots+a_{n} \cdot n$ !, where $a_{k}$ are integers, and $0 \leqslant a_{k} \leqslant k(k=1,2,3, \cdots, n)$, find the values of $a_{k}$.
45
$$
1986=a_{1}+a_{2} \cdot 2!+a_{3} \cdot 3!+\cdots+a_{6} \cdot 6!,
$$

which is
$$
1986=a_{1}+2 a_{2}+6 a_{3}+24 a_{4}+120 a_{5}+720 a_{6} .
$$

Obviously, $a_{1}$ is even, and $a_{1}=1,0$, so $a_{1}=0$. Equation (1) becomes
$$
1986=2 a_{2}+6 a_{3}+24 a_{4}+120 a_{5}+720 a_{6} .
$$

From the above equation, $a_{2}$ is a multiple of 3, and $0 \leqslant a_{2} \leqslant 2$, hence $a_{2}=0$. Similarly, we know that
$$
a_{3} \equiv 3(\bmod 4),
$$

and $0 \leqslant a_{3} \leqslant 3$, so $a_{3}=3$;
$$
a_{4} \equiv 2(\bmod 5),
$$

and $0 \leqslant a_{4} \leqslant 4$, so $a_{4}=2$;
$$
a_{5} \equiv 4(\bmod 6),
$$

and $0 \leqslant a_{5} \leqslant 5$, so $a_{5}=4$, thus $a_{6}=2$.
In summary, $a_{1}=a_{2}=0, a_{3}=3, a_{4}=2, a_{5}=4, a_{6}=2$.