$33 \cdot 54$ Let $F=0.48181 \cdots \cdots$ be an infinite repeating decimal, where 8 and 1 are the repeating digits. When $F$ is written as a simplified fraction, the denominator exceeds the numerator by
(A) 13.
(B) 14.
(C) 29.
(D) 57.
(E) 126.
(21st American High School Mathematics Examination, 1970)
$\begin{array}{l}\text { [Solution] } \begin{array}{l} F=0.4818181 \cdots \cdots=0.4+0.0818181 \cdots \cdots \\ =0.4+\frac{1}{10} \cdot 0.818181 \cdots \cdots \\ =\frac{4}{10}+\frac{1}{10} \cdot \frac{81}{99}=\frac{53}{110}, \\ \text { Since } 110-53=57 . \\ \text { Therefore, the answer is }(D)\end{array}\end{array}$