1970 AHSME Problems/Problem 10

Problem

Let $F=.48181\cdots$ 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

$\text{(A) } 13\quad \text{(B) } 14\quad \text{(C) } 29\quad \text{(D) } 57\quad \text{(E) } 126$

Solution

Multiplying by $100$ gives $100F = 48.181818...$ . Subtracting the first equation from the second gives $99F = 47.7$ , and all the other repeating parts cancel out. This gives $F = \frac{47.7}{99} = \frac{477}{990} = \frac{159}{330} = \frac{53}{110}$ . Subtracting the numerator from the denominator gives $\fbox{D}$ .

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1970 AHSME Problems/Problem 10

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