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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 (ProblemsResources)
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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