1977 AHSME Problems/Problem 28
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).
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 — contradicted by the solutions. Thus, the only remaining possibility is that the remainder is a constant .