1977 AHSME Problems/Problem 28

Problem

Solution 1

Let $r(x)$ be the remainder when $g(x^{12})$ is divided by $g(x)$. Then $r(x)$ is the unique polynomial such that \[g(x^{12}) - r(x) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1 - r(x)\] is divisible by $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1$, and $\deg r(x) < 5$.

Note that $(x - 1)(x^5 + x^4 + x^3 + x^2 + 1) = x^6 - 1$ is a multiple of $g(x)$. Also, \[g(x^{12}) - 6 = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} - 5 \\ = (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1).\] Each term is a multiple of $x^6 - 1$. For example, \[x^{60} - 1 = (x^6 - 1)(x^{54} + x^{48} + \dots + x^6 + 1).\] Hence, $g(x^{12}) - 6$ is a multiple of $x^6 - 1$, which means that $g(x^{12}) - 6$ is a multiple of $g(x)$. Therefore, the remainder is $\boxed{6}$. The answer is (A).

Solution 2

We express the quotient and remainder as follows. \[g(x^{12}) = Q(x) g(x) + R(x)\] Note that the solutions to $g(x)$ correspond to the 6th roots of unity, excluding $1$. Hence, we have $x^6 = 1$, allowing us to set: \[g(x^{12}) = 6\] \[g(x) = 0\] We have $5$ values of $x$ that return $R(x) = 6$. However, $g(x)$ is quintic, implying the remainder is of degree $4$ — contradicted by the $5$ solutions. Thus, the only remaining possibility is that the remainder is a constant $\boxed{6}$.

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