2008 iTest Problems/Problem 80

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
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100

Page

Toolbox

Search

2008 iTest Problems/Problem 80

Contents

Problem

Solutions

Solution 1

Solution 2

See Also