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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) = 0</cmath>
 
<cmath>g(x) = 0</cmath>
 
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>.
 
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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== See also ==
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{{AHSME box | year = 1977 | num-b = 27 | after = 29}}

Revision as of 19:21, 31 October 2021

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 $\boxed{6}$.

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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