Outrageous Quotient And Remainder

Let $R$ and $S$ be the next polinomials:

$R(x) = x(x^{2006} - 1)$ $S(x) = x^{2} + x +1$

We use this polinomials to solve this problem: $P = 2008^{2007} - 2008 = 2008 \cdot (2008^{2006} -1)$ so $P = R(2008)$ $Q = 2008^{2} + 2009 = 2008^{2} + 2008 +1$ so $Q = S(2008)$

We know that for every natural number $n$ ,

$x^{n} - 1 = (x - 1) \cdot (x^{n-1} + x^{n-2} + \cdots + x^{2} + x + 1)$

Applying this relation to $R(x)$ we have,

$\begin{aligned} R(X) &= x \cdot (x - 1) \cdot ( x^{2005} +x^{2004} + \cdots + x^{2} + x + 1) \\ &= (x - 1) \cdot ( x^{2006} + x^{2005} + \cdots + x^{3} + x^{2} + x) \\ &= (x - 1) \cdot [(x^{2004} + x^{2001} + \cdots + x^{6} + x^{3} + 1) \cdot (x^{2} + x + 1) - 1] \\ &\equiv (1 - x) \pmod{S(x)} \\ &\equiv (x^{2} + 2) \pmod{S(x)} \\ \end{aligned}$

Then

$\begin{aligned} P &= R(2008) \\ &\equiv (2008^{2} + 2)\pmod{S(2008)} \\ &\equiv \color{#D61F06} {4032066}\pmod{Q} \\ \end{aligned}$

So, the answer is $\fbox{ 4032066 }$

notice that P is in the form $x^{2007}$ -x and Q is in the form $x^{2}$ +x+1

let f(x)= $x^{2007}$ -x g(x)= $x^{2}$ +x+1

f(x)=g(x)*q(x)+R

Let g(x) be 0.

Then f(x)=R

$x^{2}$ +x+1=0

$x^{3}$ =1

putting the value in f(x)

f(x)= $x^2$ +2 =2008*2008+2 =4032066

R=4032066

using multiple subtractions , notice that a mod b = (a-b) mod b finally got the answer ^_^ , great problems