1989 AIME Problems/Problem 3
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 .
(Note: Any repeating sequence of digits that looks like can be written as , where represents an digit number.)
To get rid of repeating decimals, we multiply the equation by 1000. We get We subtract the original equation from the second to get We simplify to Since is an integer, because is relatively prime to , and d25 is divisible by but not . The only odd number that yields a single digit and 25 at the end of the three digit number is , so the answer is .
Similar to Solution 2, we start off by writing that .Then we subtract this from the original equation to get:
Since n is an integer, we have that .
Since is prime, we can apply Euclid's Lemma (which states that if is a prime and if and are integers and if , then or ) to realize that , since . Then we can expand as . Since , by Euclid, we can arrive at . From this we know that . (This is true because )
Write out these equations:
Thus divides 25 and 30. The only solution for this under 1000 is .
Video Solution by OmegaLearn
|1989 AIME (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|