1970 AHSME Problems/Problem 10

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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 $\fbox{D}$.

See also

1970 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
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