Difference between revisions of "1989 AIME Problems/Problem 3"
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(Note: Any repeating sequence of <math>n</math> digits that looks like <math>0.a_1a_2a_3...a_{n-1}a_na_1a_2...</math> can be written as <math>\frac{a_1a_2...a_n}{10^n-1}</math>, where <math>a_1a_2...a_n</math> represents an <math>n</math> digit number.) | (Note: Any repeating sequence of <math>n</math> digits that looks like <math>0.a_1a_2a_3...a_{n-1}a_na_1a_2...</math> can be written as <math>\frac{a_1a_2...a_n}{10^n-1}</math>, where <math>a_1a_2...a_n</math> represents an <math>n</math> digit number.) | ||
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+ | == Solution 2 == | ||
+ | To get rid of repeating decimals, we multiply the equation by 1000. We get <math>\frac{1000n}{810} = d25.d25d25...</math> We subtract the original equation from the second to get <math>\frac{999n}{810}=d25</math> We simplify to <math>\frac{37n}{30} = d25</math> Since <math>\frac{37n}{30}</math> is an integer, <math>n=(30)(5)(2k+1)</math> because <math>37</math> is relatively prime to <math>30</math>, and d25 is divisible by <math>5</math> but not <math>10</math>. The only odd number that yields a single digit <math>d</math> and 25 at the end of the three digit number is <math>k=2</math>, so the answer is <math>750</math>. | ||
== See also == | == See also == |
Revision as of 07:04, 27 July 2017
Contents
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 .
(Note: Any repeating sequence of digits that looks like can be written as , where represents an digit number.)
Solution 2
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 .
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.