Difference between revisions of "2008 iTest Problems/Problem 80"

Latest revision as of 18:49, 23 November 2018

Problem

Let

$p(x) = x^{2008} + x^{2007} + x^{2006} + \cdots + x + 1,$

and let $r(x)$ be the polynomial remainder when $p(x)$ is divided by $x^4+x^3+2x^2+x+1$ . Find the remainder when $|r(2008)|$ is divided by $1000$ .

Solutions

Solution 1

$x^4+x^3+2x^2+x+1 = (x^2 + 1)(x^2 + x + 1)$ . We apply the polynomial generalization of the Chinese Remainder Theorem.

Indeed,

$p(x) = (x^{2008} + x^{2007} + x^{2006}) + \cdots + (x^4 + x^3 + x^2) + x + 1 \equiv x+1 \pmod{x^2 + x + 1}$

since $x^{n+2} + x_{n+1} + x^{n} = x^{n-2}(x^2 + x + 1) \equiv 0 \pmod{x^2 + x + 1}$ . Also,

$p(x) = (x^{2008} + x^{2006}) + (x^{2007} + x^{2005}) + \cdots + (x^4 + x^2) + (x^3 + x) + 1 \equiv 1 \pmod{x^2 + 1}$

using similar reasoning. Hence $p(x) \equiv x+1 \pmod{x^2 + x + 1}, p(x) \equiv 1 \pmod{x^2 + 1}$ , and by CRT we have $p(x) \equiv -x^2 \pmod{x^4+x^3+2x^2+x+1}$ .

Then $|r(2008)| \equiv 2008^2 \equiv \boxed{64} \pmod{1000}$ .

Solution 2

The given information can be represented as $x^{2008} + x^{2007} + \cdots + x^2 + x + 1 = q(x) \cdot (x^4 + x^3 + 2x^2 + x + 1) + r(x)$ where $r(x)$ is the remainder and $q(x)$ is the quotient. Multiplying both sides by $x-1$ would make computation easier; doing so results in $\begin{align*} x^{2009} - 1 &= q(x) \cdot (x^4 + x^3 + 2x^2 + x + 1)(x-1) + r(x) \cdot (x-1) \\ &= q(x) \cdot (x^2 + 1)(x^2 + x + 1)(x-1) + r(x) \cdot (x-1) \end{align*}$ Now plug in values of $x$ not equal to 1 that make $q(x) \cdot (x^2 + 1)(x^2 + x + 1)(x-1) = 0$ . By the Zero Product Property, the values of $x$ that make the expression equal to 0 are $\pm i$ and $\tfrac{-1 \pm i\sqrt{3}}{2}$ .

By plugging in $\pm i$ and $\tfrac{-1 \pm i\sqrt{3}}{2}$ into the equation and solving for the remainder function, we have $r(\pm i) = 1, r(\tfrac{-1 + i\sqrt{3}}{2}) = \tfrac{1 + i\sqrt{3}}{2}, r(\tfrac{-1 - i\sqrt{3}}{2}) = \tfrac{1 - i\sqrt{3}}{2}$ . Since the remainder function's degree is at most 3 and we know four points, we can construct the unique remainder function.

Let $r(x) = ax^3 + bx^2 + cx + d$ . Plugging in $i$ results in $-ai - b + ci + d = 1$ , and plugging in $-i$ results in $ai - b - ci + d = 1$ . Thus, $a = c$ and $d-b = 1$ . Plugging in $\tfrac{-1 + i\sqrt{3}}{2}$ results in $a + \tfrac{-1 - i\sqrt{3}}{2}b + \tfrac{-1 + i\sqrt{3}}{2}a + d = \tfrac{1 + i\sqrt{3}}{2}$ . Plugging in $\tfrac{-1 - i\sqrt{3}}{2}$ results in $a + \tfrac{-1 + i\sqrt{3}}{2} + \tfrac{-1 - i\sqrt{3}}{2}a + d = \tfrac{1 - i\sqrt{3}}{2}$ . Thus, $-b + a = 1$ and $\tfrac12 a - \tfrac12 b + d = \tfrac12$ , so $\begin{align*} a - b + 2d &= 1 \\ 1 + 2d &= 1 \\ d &= 0 \end{align*}$ Substituting $d = 0$ results in $b = -1$ and $a = c = 0$ . Therefore, $r(x) = -x^2$ , so $|r(2008)| = 4032064$ , and the remainder when divided by 1000 is $\boxed{64}$ .

2008 iTest (Problems)
Preceded by: Problem 79	Followed by: Problem 81
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@@ Line 5: / Line 5: @@
 the remainder when <math>|r(2008)|</math> is divided by <math>1000</math>.
-== Solution ==
+== Solutions ==
+===Solution 1===
 <math>x^4+x^3+2x^2+x+1 = (x^2 + 1)(x^2 + x + 1)</math>. We apply the polynomial generalization of the [[Chinese Remainder Theorem]].
@@ Line 18: / Line 21: @@
 using similar reasoning. Hence <math>p(x) \equiv x+1 \pmod{x^2 + x + 1}, p(x) \equiv 1 \pmod{x^2 + 1}</math>, and by CRT we have <math>p(x) \equiv -x^2 \pmod{x^4+x^3+2x^2+x+1}</math>.
-Then <math>|r(2008)| \equiv 2008^2 \equiv 64 \pmod{1000}</math>.
+Then <math>|r(2008)| \equiv 2008^2 \equiv \boxed{64} \pmod{1000}</math>.
+===Solution 2===
+The given information can be represented as
+<cmath>x^{2008} + x^{2007} + \cdots + x^2 + x + 1 = q(x) \cdot (x^4 + x^3 + 2x^2 + x + 1) + r(x)</cmath>
+where <math>r(x)</math> is the remainder and <math>q(x)</math> is the quotient. Multiplying both sides by <math>x-1</math> would make computation easier; doing so results in
+<cmath>\begin{align*}
+x^{2009} - 1 &= q(x) \cdot (x^4 + x^3 + 2x^2 + x + 1)(x-1) + r(x) \cdot (x-1) \\
+&= q(x) \cdot (x^2 + 1)(x^2 + x + 1)(x-1) + r(x) \cdot (x-1)
+\end{align*}</cmath>
+Now plug in values of <math>x</math> not equal to 1 that make <math>q(x) \cdot (x^2 + 1)(x^2 + x + 1)(x-1) = 0</math>. By the Zero Product Property, the values of <math>x</math> that make the expression equal to 0 are <math>\pm i</math> and <math>\tfrac{-1 \pm i\sqrt{3}}{2}</math>.
+<br>
+By plugging in <math>\pm i</math> and <math>\tfrac{-1 \pm i\sqrt{3}}{2}</math> into the equation and solving for the remainder function, we have <math>r(\pm i) = 1, r(\tfrac{-1 + i\sqrt{3}}{2}) = \tfrac{1 + i\sqrt{3}}{2}, r(\tfrac{-1 - i\sqrt{3}}{2}) = \tfrac{1 - i\sqrt{3}}{2}</math>. Since the remainder function's degree is at most 3 and we know four points, we can construct the unique remainder function.
+<br>
+Let <math>r(x) = ax^3 + bx^2 + cx + d</math>. Plugging in <math>i</math> results in <math>-ai - b + ci + d = 1</math>, and plugging in <math>-i</math> results in <math>ai - b - ci + d = 1</math>. Thus, <math>a = c</math> and <math>d-b = 1</math>. Plugging in <math>\tfrac{-1 + i\sqrt{3}}{2}</math> results in <math>a + \tfrac{-1 - i\sqrt{3}}{2}b + \tfrac{-1 + i\sqrt{3}}{2}a + d = \tfrac{1 + i\sqrt{3}}{2}</math>. Plugging in <math>\tfrac{-1 - i\sqrt{3}}{2}</math> results in <math>a + \tfrac{-1 + i\sqrt{3}}{2} + \tfrac{-1 - i\sqrt{3}}{2}a + d = \tfrac{1 - i\sqrt{3}}{2}</math>. Thus, <math>-b + a = 1</math> and <math>\tfrac12 a - \tfrac12 b + d = \tfrac12</math>, so
+<cmath>\begin{align*}
+a - b + 2d &= 1 \\
++ 2d &= 1 \\
+d &= 0
+\end{align*}</cmath>
+Substituting <math>d = 0</math> results in <math>b = -1</math> and <math>a = c = 0</math>. Therefore, <math>r(x) = -x^2</math>, so <math>|r(2008)| = 4032064</math>, and the remainder when divided by 1000 is <math>\boxed{64}</math>.
+== See Also ==
+{{2008 iTest box|num-b=79|num-a=81}}
-== See also ==
 [[Category:Intermediate Algebra Problems]][[Category:Intermediate Number Theory Problems]]

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