Difference between revisions of "1989 AIME Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], <math> | + | Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], <math>0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}</math>. Thus <math>\frac{n}{810} = \frac{100d + 25}{999}</math> so <math>n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}</math>. Since 750 and 37 are [[relatively prime]], <math>4d + 1</math> must be [[divisible]] by 37, and the only digit for which this is possible is <math>d = 9</math>. Thus <math>4d + 1 = 37</math> and <math>n = 750</math>. |
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 18:15, 4 July 2013
Problem
Suppose 
is a positive integer and 
is a single digit in base 10. Find if
Solution
Repeating decimals represent rational numbers. To figure out which rational number, we sum an infinite geometric series, 
. Thus 
so 
. Since 750 and 37 are relatively prime, 
must be divisible by 37, and the only digit for which this is possible is 
. Thus 
and 
.
See also
| 1989 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
