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

Revision as of 22:05, 10 February 2017

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).

@@ Line 4: / Line 4: @@
 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>
+<cmath>
-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>
+</cmath>
 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>
 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).

Page

Toolbox

Search

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

Revision as of 22:05, 10 February 2017