1970 AMC 12 Problem 10

Attempt Problem 10 of the 1970 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1970 AMC 12 solutions, or check the answer key.

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10.

Let

F=0.4818181\ldots

be an infinite repeating decimal with the digits

8

and

1

repeating. When

F

is written as a fraction in lowest terms, the denominator exceeds the numerator by:

13

14

29

57

126

Separate the initial

0.4

from the repeating tail

The tail is

0.081(1+0.01+0.01^2+\cdots)

We have

\begin{aligned} F&=0.4+0.081(1+0.01+\cdots)\\ &=\frac25+\frac{0.081}{0.99}\\ &=\frac25+\frac9{110}\\ &=\frac{53}{110}. \end{aligned}

The denominator exceeds the numerator by

110-53=57.

Therefore, the correct answer is D.

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