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1989 AIME Problems/Problem 3

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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, $\displaystyle 0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}$ . 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 $n = 750$ .

See also

1989 AIME (Problems • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Category: Intermediate Algebra Problems

Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.

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