Question

$P={2008}^{2007};Q={2008}^2+2009$. The remainder when P is divided by $Q$ is

• A. $4031964$
• B. $4032066$
• C. $4158972$
• D. $40682896$

Answer 1

The correct option is B $4032066$

Given,

$P={2008}^{2007}-2008$

$Q={2008}^2-2009$

let us assume $x=2008$

$\therefore$ $P=x^{2007}-x=(x-1)(x+x^2+x^3+x^4+......x^{2006})$

$=(x-1)(x(1+x)+x^3(1+x+x^2)+......+x^{2001}(1+x+x^2)+x^{2004}(1+x+x^2))$

$=x(x^2-1)+x^3(x-1)(1+x+x^2)+......+x^{2001}(x-1)(1+x+x^2)+x^{2004}(x-1)(1+x+x^2)$

Now, $1+x+x^2=1+2008+{2008}^2=2009+{2008}^2=Q$

$\therefore$ $\frac{P}{Q}=\frac{x(x^2-1)+x^3(x-1)(1+x+x^2)+......+x^{2004}(x-1)(1+x+x^2)}{1+x+x^2}$

So we can clearly see that the only term remaining which will be remainder is

$\frac{P}{Q}=\frac{x^3-x}{1+x+x^2}=(x-1)+\left(\frac{1-x}{1+x+x^2}\right)$

$\therefore$ the remainder of $\left(\frac{P}{Q}\right)$ is $1-x=1-2008=-2007$

But the remainder cannot be negative, So the actual remainder is

$Q-2007={\left(2008\right)}^2+2009-2007$

$={\left(2008\right)}^2+2$

$=4032066$

Answer : Option B