Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.3 Junior Find the answer

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

Pick one

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 $\fbox{D}$ .

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