2019 AIME I Problems/Problem 8
The 2019 AIME I takes place on March 13, 2019.
Contents
Problem 8
Let be a real number such that . Then where and are relatively prime positive integers. Find .
Solution
We can substitute . Since we know that , we can do some simplification.
This yields . From this, we can substitute again to get some cancellation through binomials. If we let , we can simplify the equation to . After using binomial theorem, this simplifies to . If we use the quadratic theorem, we obtain that , so . By plugging z into (which is equal to , we can either use binomial theorem or sum of cubes to simplify, and we end up with 13/54. Therefore, the answer is .
eric2020, inspired by Tommy2002
Solution 2
First, for simplicity, let and . Note that . We then bash the rest of the problem out. Take the tenth power of this expression and get . Note that we also have . So, it suffices to compute . Let . We have from cubing that or . Next, using , we get or . Solving gives or . Clearly is extraneous, so . Now note that , and . Thus we finally get , giving .
-Emathmaster
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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