$33 \cdot 66$ In a certain base of numeration, the square of 24 is 554, the base is
(A) 6.
(B) 8.
(C) 12.
(D) 14.
(E) 16.
(24th American High School Mathematics Examination, 1973)
[Solution] Let the base be $b$, and $b>5$.
Since $\left(24_{b}\right)^{2}=(2 b+4)^{2}=4 b^{2}+16 b+16$
and $\quad 554=5 b^{2}+5 b+4$,
we have $5 b^{2}+5 b+4=4 b^{2}+16 b+16$.
Solving this, we get $b=-1$ or 12, discarding the negative value.
Therefore, the answer is $(C)$.