Difference between revisions of "1977 AHSME Problems/Problem 28"

Revision as of 12:13, 8 June 2022

Problem

Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$ ?

$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$

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 $\boxed{6}$ .

Solution 3

We can use the Chinese remainder theorem over $\mathbb{Q}[x]$ . Since $g(x)=(x+1)(x^2-x+1)(x^2+x+1)$ , $\mathbb{Q}[x]/g\cong \mathbb{Q}[x]/(x+1)\times \mathbb{Q}[x]/(x^2-x+1)\times \mathbb{Q}[x]/(x^2+x+1)$ . This means that if we can find the remainder of $g(x^12)$ modulo $x+1,x^2-x+1,x^2+x+1$ , we can reconstruct the remainder modulo $g$ . We can further use that each factor is irreducible and that if $p(x)$ is an irreducible polynomial over $\mathbb{Q}$ with root $\alpha$ , $\mathbb{Q}[x]/p\cong \mathbb{Q}(\alpha)$ so to evaluate the remainders of $g(x^{12})$ , we just need to evaluate it on one of the roots of the irreducible factors. The first factor has root $-1$ , 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 $g(x)(x-1)=x^6-1$ ). Evaluating $g(x^{12})$ on each of these values yields $g(1)=6$ so the remainder is $6$ on each factor on the right of the isomorphism. Hence, by the Chinese remainder theorem, the remainder modulo $g$ must be $\boxed{6}$ as well.

@@ Line 33: / Line 33: @@
 ===Solution 3===
-We can use the Chinese remainder theorem over <math>\mathbb{Q}[x]</math>. Since <math>g(x)=(x+1)(x^2-x+1)(x^2+x+1)</math>, <math>\mathbb{Q}[x]/g\cong \mathbb{Q}[x]/(x+1)\times \mathbb{Q}[x]/(x^2-x+1)\times \mathbb{Q}[x]/(x^2+x+1)</math>. This means that if we can find the remainder of <math>g(x^12)</math> modulo <math>x+1,x^2-x+1,x^2+x+1</math>, we can reconstruct the remainder modulo <math>g</math>. We can further use that each factor is irreducible and that if <math>p(x)</math> is an irreducible polynomial over <math>\mathbb{Q}</math> with root <math>\alpha</math>, <math>\mathbb{Q}[x]/p\cong \mathbb{Q}(\alpha)</math> so to evaluate the remainders of <math>g(x^12)</math>, we just need to evaluate it on one of the roots of the irreducible factors. The first factor has root <math>-1</math>, 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 <math>g(x)(x-1)=x^6-1</math>). Evaluating <math>g(x^12)</math> on each of these values yields <math>g(1)=6</math> so the remainder is <math>6</math> on each factor on the right of the isomorphism. Hence, by the Chinese remainder theorem, the remainder modulo <math>g</math> must be <math>\boxed{6}</math> as well.
+We can use the Chinese remainder theorem over <math>\mathbb{Q}[x]</math>. Since <math>g(x)=(x+1)(x^2-x+1)(x^2+x+1)</math>, <math>\mathbb{Q}[x]/g\cong \mathbb{Q}[x]/(x+1)\times \mathbb{Q}[x]/(x^2-x+1)\times \mathbb{Q}[x]/(x^2+x+1)</math>. This means that if we can find the remainder of <math>g(x^12)</math> modulo <math>x+1,x^2-x+1,x^2+x+1</math>, we can reconstruct the remainder modulo <math>g</math>. We can further use that each factor is irreducible and that if <math>p(x)</math> is an irreducible polynomial over <math>\mathbb{Q}</math> with root <math>\alpha</math>, <math>\mathbb{Q}[x]/p\cong \mathbb{Q}(\alpha)</math> so to evaluate the remainders of <math>g(x^{12})</math>, we just need to evaluate it on one of the roots of the irreducible factors. The first factor has root <math>-1</math>, 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 <math>g(x)(x-1)=x^6-1</math>). Evaluating <math>g(x^{12})</math> on each of these values yields <math>g(1)=6</math> so the remainder is <math>6</math> on each factor on the right of the isomorphism. Hence, by the Chinese remainder theorem, the remainder modulo <math>g</math> must be <math>\boxed{6}</math> as well.
 == See also ==
 {{AHSME box | year = 1977 | num-b = 27 | after = 29}}

1977 AHSME (Problems • Answer Key • Resources)
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