$34 \cdot 6$ number 695 is written as a polynomial in factorials: $695=a_{1}+a_{2} \cdot 2!+$ $a_{3} \cdot 3!+\cdots+a_{n} \cdot n!$, where $a_{1}, a_{2}, \cdots, a_{k}$ are integers, and $0 \leqslant a_{k} \leqslant k$, then $a_{4}$ equals
(A) 0 .
(B) 1 .
(C) 2 .
(D) 3 .
(E) 4 .
(12th American High School Mathematics Examination, 1961)
[Solution] From $6!>695$, we know we can set
$$
695=a_{1}+a_{2} \cdot 2!+a_{3} \cdot 3!+a_{4} \cdot 4!+a_{5} \cdot 5!,
$$

i.e., $695=a_{1}+2 a_{2}+6 a_{3}+24 a_{4}+120 a_{5}$, where $0 \leqslant a_{k} \leqslant k$.
From the above, $a_{5}=5$ (otherwise the sum cannot reach 695).

Also, when $a_{5}=5$, $a_{4} \neq 4$ (note that $5 \cdot 120+4 \cdot 24>695$),
and $a_{4} \geqslant 3$ (otherwise $a_{4} \cdot 4!+3 \cdot 6+2 \cdot 2+1<95$, if $a_{4}<3$).
Therefore, $a_{4}=3$.
If $a_{5}=5, a_{4}=3$, then $a_{3}=3, a_{2}=2, a_{1}=1$.
Thus, the answer is $(D)$.