1989 AIME Problems/Problem 3

Problem

Suppose $n$ is a positive integer and $d$ is a single digit in base 10. Find $n$ if

$\frac{n}{810}=0.d25d25d25\ldots$

Solution

Repeating decimals represent rational numbers. To figure out which rational number, we sum an infinite geometric series, $0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}$ . Thus $\frac{n}{810} = \frac{100d + 25}{999}$ so $n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}$ . Since 750 and 37 are relatively prime, $4d + 1$ must be divisible by 37, and the only digit for which this is possible is $d = 9$ . Thus $4d + 1 = 37$ and $n = 750$ .

(Note: Any repeating sequence of $n$ digits that looks like $0.a_1a_2a_3...a_{n-1}a_na_1a_2...$ can be written as $\frac{a_1a_2...a_n}{10^n-1}$ , where $a_1a_2...a_n$ represents an $n$ digit number.)

1989 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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1989 AIME Problems/Problem 3

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