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> | + | <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). | ||
− | + | </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 17:10, 3 December 2016
Let be the remainder when is divided by . Then is the unique polynomial such that is divisible by , and .
Note that is a multiple of . 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 . For example, Hence, is a multiple of , which means that is a multiple of . Therefore, the remainder is . The answer is (A).