Difference between revisions of "2023 AIME II Problems/Problem 8"
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Because the answer must be a positive integer, it is just equal to the modulus of the product. Define <math>z_n = \left(\textrm{cis }\frac{2n\pi}{7}\right)^3 + \textrm{cis }\frac{2n\pi}{7} + 1</math>. | Because the answer must be a positive integer, it is just equal to the modulus of the product. Define <math>z_n = \left(\textrm{cis }\frac{2n\pi}{7}\right)^3 + \textrm{cis }\frac{2n\pi}{7} + 1</math>. | ||
+ | |||
+ | Then, our product is equal to | ||
+ | |||
+ | <cmath>z_0z_1z_2z_3z_4z_5z_6.</cmath> | ||
== See also == | == See also == | ||
{{AIME box|year=2023|num-b=7|num-a=9|n=II}} | {{AIME box|year=2023|num-b=7|num-a=9|n=II}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:16, 16 February 2023
Solution 1
For any , we have, The second and the fifth equalities follow from the property that .
Therefore,
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Moduli)
Because the answer must be a positive integer, it is just equal to the modulus of the product. Define .
Then, our product is equal to
See also
2023 AIME II (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.