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

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==Problem==
 
==Problem==
  
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Let <math>g(x)=x^5+x^4+x^3+x^2+x+1</math>. What is the remainder when the polynomial <math>g(x^{12})</math> is divided by the polynomial <math>g(x)</math>?
  
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<math>\textbf{(A) }6\qquad
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\textbf{(B) }5-x\qquad
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\textbf{(C) }4-x+x^2\qquad
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\textbf{(D) }3-x+x^2-x^3\qquad \\
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\textbf{(E) }2-x+x^2-x^3+x^4</math> 
  
  
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<cmath>g(x^{12}) = 6</cmath>
 
<cmath>g(x^{12}) = 6</cmath>
 
<cmath>g(x) = 0</cmath>
 
<cmath>g(x) = 0</cmath>
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We have <math>5</math> values of <math>x</math> that return <math>R(x) = 6</math>. However, <math>g(x)</math> is quintic, implying the remainder is of degree <math>4</math> — contradicted by the <math>5</math> solutions. Thus, the only remaining possibility is that the remainder is a constant <math>\boxed{6}</math>.
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We have <math>5</math> values of <math>x</math> that return <math>R(x) = 6</math>. However, <math>g(x)</math> is quintic, implying the remainder is of degree at most <math>4</math>. Since there are <math>5</math> solutions, the only possibility is that the remainder is a constant <math>\boxed{6}</math>.
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===Solution 3===
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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.
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== See also ==
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{{AHSME box | year = 1977 | num-b = 27 | after = 29}}
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[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 10:56, 21 April 2024

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 at most $4$. Since there are $5$ solutions, the only 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.

See also

1977 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
29
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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