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

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Note that <math>(x - 1)(x^5 + x^4 + x^3 + x^2 + 1) = x^6 - 1</math> is a multiple of <math>g(x)</math>. Also,
 
Note that <math>(x - 1)(x^5 + x^4 + x^3 + x^2 + 1) = x^6 - 1</math> is a multiple of <math>g(x)</math>. Also,
<math>\begin{align*}
+
<math>
 
g(x^{12}) - 6 &= x^{60} + x^{48} + x^{36} + x^{24} + x^{12} - 5 \\
 
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).
 
&= (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1).
\end{align*}</math>
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</math>
 
Each term is a multiple of <math>x^6 - 1</math>. For example,
 
Each term is a multiple of <math>x^6 - 1</math>. For example,
 
<cmath>x^{60} - 1 = (x^6 - 1)(x^{54} + x^{48} + \dots + x^6 + 1).</cmath>
 
<cmath>x^{60} - 1 = (x^6 - 1)(x^{54} + x^{48} + \dots + x^6 + 1).</cmath>
 
Hence, <math>g(x^{12}) - 6</math> is a multiple of <math>x^6 - 1</math>, which means that <math>g(x^{12}) - 6</math> is a multiple of <math>g(x)</math>. Therefore, the remainder is <math>\boxed{6}</math>. The answer is (A).
 
Hence, <math>g(x^{12}) - 6</math> is a multiple of <math>x^6 - 1</math>, which means that <math>g(x^{12}) - 6</math> is a multiple of <math>g(x)</math>. Therefore, the remainder is <math>\boxed{6}</math>. The answer is (A).

Revision as of 18:10, 3 December 2016

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, $g(x^{12}) - 6 &= x^{60} + x^{48} + x^{36} + x^{24} + x^{12} - 5 \\ &= (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1).$ (Error compiling LaTeX. Unknown error_msg) 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).