$33 \cdot 62$  Each number in this problem is expressed in base $b$. Given that the product (12)(15)(16) $=3146$, and let $s=12+15+16$, then $s$ is $(b$ base $)$
(A) 43 .
(B) 44 .
(C) 45 .
(D) 46 .
(E) 47 .
(18th American High School Mathematics Examination, 1967)
[Solution] In base $b$, $12_{b}, 15_{b}, 16_{b}, 3146_{b}$ are respectively
$$
b^{1}+2, b^{1}+5, b^{1}+6, 3 b^{3}+b^{2}+4 b+6 \text {. }
$$

Assuming $(b+2)(b+5)(b+6)=3 b^{3}+b^{2}+4 b+6$, simplifying gives $\quad b^{3}-6 b^{2}-24 b-27=0$.
Solving for $b$ yields $b=9$ as the only real solution.
Furthermore, $s=12_{b}+15_{b}+16_{b}=(b+2)+(b+5)+(b+6)$ $=3 b+13=4 b+4=44_{b}$.
Therefore, the answer is $(B)$.