1977 AHSME Problems/Problem 28
Problem
Let . What is the remainder when the polynomial is divided by the polynomial ?
Solution 1
Let be the remainder when is divided by . Then is the unique polynomial such that is divisible by , and .
Note that is a multiple of . Also, Each term is a multiple of . For example, Hence, is a multiple of , which means that is a multiple of . Therefore, the remainder is . The answer is (A).
Solution 2
We express the quotient and remainder as follows. Note that the solutions to correspond to the 6th roots of unity, excluding . Hence, we have , allowing us to set: We have values of that return . However, is quintic, implying the remainder is of degree at most . Since there are solutions, the only possibility is that the remainder is a constant .
Solution 3
We can use the Chinese remainder theorem over Since , This means that if we can find the remainder of modulo , we can reconstruct the remainder modulo We can further use that each factor is irreducible and that if is an irreducible polynomial over with root , so to evaluate the remainders of , we just need to evaluate it on one of the roots of the irreducible factors. The first factor has root , the second has roots the primitive sixth roots of unity, and the third as roots the primitive cube roots of unity (this is easily seen as ). Evaluating on each of these values yields so the remainder is on each factor on the right of the isomorphism. Hence, by the Chinese remainder theorem, the remainder modulo must be as well.
See also
1977 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by 29 | |
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