$33 \cdot 59$ number 10! (in decimal) when written in base 12, ends with exactly $k$ zeros. Then the value of $k$ is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(E) 5.
(21st American High School Mathematics Examination, 1970)
[Solution] In decimal,
$$
\begin{array}{l}
10!=1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10=12^{4} \cdot 5^{2} \cdot 7 \text {. } \\
\text { and } \quad 5^{2} \cdot 7=175=1 \cdot 12^{2}+2 \cdot 12+7 \text {, } \\
\text { thus }\left(5^{2} \cdot 7 \cdot 12^{4}\right)_{10}=\left(127 \cdot 10^{4}\right)_{12}=1270000_{12} \text {. } \\
\text { Therefore, the answer is }(D) \text {. } \\
\end{array}
$$