2021 AMC 12B Problems/Problem 20
Contents
Problem
Let and be the unique polynomials such thatand the degree of is less than What is
Solution 1
Note that so if is the remainder when dividing by , Now, So , and The answer is
Solution 1b (More Thorough Version of 1)
Instead of dealing with a nasty , we can instead deal with the nice , as is a factor of . Then, we try to see what is. Of course, we will need a , getting . Then, we've gotta get rid of the term, so we add a , to get . This pattern continues, until we add a to get rid of , and end up with . We can't add anything more to get rid of the , so our factor is . Then, to get rid of the , we must have a remainder of , and to get the we have to also have a in the remainder. So, our product is Then, our remainder is . The remainder when dividing by must be the same when dividing by , modulo . So, we have that , or . This corresponds to answer choice . ~rocketsri
Solution 2 (Complex numbers)
One thing to note is that takes the form of for some constants A and B. Note that the roots of are part of the solutions of They can be easily solved with roots of unity: Obviously the right two solutions are the roots of We substitute into the original equation, and becomes 0. Using De Moivre's theorem, we get: Expanding into rectangular complex number form: Comparing the real and imaginary parts, we get: The answer is . ~Jamess2022(burntTacos;-;)
Video Solution by OmegaLearn (Using Modular Arithmetic and Meta-solving)
~ pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.