Question

$P = {2008^{2007}} - 2008$ and $Q = {2008^2} + 2009$, the remainder when $P$ is divided by $Q$ is:
A). $4031964$
B). $4032066$
C). $4158972$
D). $40682896$

Answer 1

Here, in the given question, the values of $P,Q$ are given and we are asked to find the remainder when $P$ is divided by $Q$. To solve this question, we have to divide $P$ by $Q$. Clearly, for the given values of $P,Q$, we cannot use a long division method here. So we will try to factorize $P$ and continue to solve to get the desired answer.

Complete step-by-step solution:
Given, $P = {2008^{2007}} - 2008$ and,
$Q = {2008^2} + 2009$
In order to factorize $P$,
Let us assume $x = 2008$
Therefore, $P = {x^{2007}} - x$ and,
$Q = {x^2} + x + 1$

\Rightarrow P = \left( {x - 1} \right)\left( {x + {x^2} + {x^3} + {x^4} + \ldots {x^{2006}}} \right)$$ We have $$2006$$ terms in the bracket, taking common $${x^2} + x + 1$$ from every next three terms, we will be left with first two individual terms $$P = \left( {x - 1} \right)\left[ {x\left( {1 + x} \right) + {x^3}\left( {1 + x + {x^2}} \right) + \ldots \ldots \ldots + {x^{2001}}\left( {1 + x + {x^2}} \right) + {x^{2004}}\left( {1 + x + {x^2}} \right)} \right]$$ Now, we know$${x^2} + x + 1 = Q$$, then, $$P = \left( {x - 1} \right)\left[ {x\left( {1 + x} \right) + {x^3}\left( Q \right) + \ldots \ldots \ldots + {x^{2001}}\left( Q \right) + {x^{2004}}\left( Q \right)} \right]$$ Taking $$Q$$ common from each term in the bracket except for first term, we get, $$P = x\left( {x - 1} \right)\left( {x + 1} \right) + Q\left( {x - 1} \right)\left[ {{x^3} + \ldots \ldots \ldots + {x^{2001}} + {x^{2004}}} \right]$$ Simplifying it, we get, $$P = \left( {{x^3} - x} \right) + Q\left( {x - 1} \right)\left[ {{x^3} + \ldots \ldots \ldots + {x^{2001}} + {x^{2004}}} \right]$$ $${x^3} - x$$ can also be written as $$\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + \left( {1 - x} \right)$$ $$P = \left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + \left( {1 - x} \right) + Q\left( {x - 1} \right)\left[ {{x^3} + \ldots \ldots \ldots + {x^{2001}} + {x^{2004}}} \right]$$ $$ \Rightarrow P = \left( {x - 1} \right)\left( Q \right) + \left( {1 - x} \right) + Q\left( {x - 1} \right)\left[ {{x^3} + \ldots \ldots \ldots + {x^{2001}} + {x^{2004}}} \right]$$ Taking $$Q\left( {x - 1} \right)$$ common from first and last term, we get, $$P = \left( {1 - x} \right) + Q\left( {x - 1} \right)\left[ {1 + {x^3} + \ldots \ldots \ldots + {x^{2001}} + {x^{2004}}} \right]$$ Clearly, if we divide $$P$$ by $$Q$$, we will get $$\left( {1 - x} \right)$$ as remainder $$\left( {1 - x} \right) = 1 - 2008 \\\ \left( {1 - x} \right) = - 2007 $$ The remainder cannot be negative, so our remainder will turn into $$Q - 2007$$ $$ = {2008^2} + 2009 - 2007$$ $$ = 4032064 + 2 \\\ = 4032066 $$ Hence, given the values of $$P,Q$$, when $$P$$ is divided by $$Q$$, the remainder is $$4032066$$ **Hence the answer is B $$4032066$$ is the correct answer.** **Note:** It is important to note here that the remainder cannot be negative. The reason for this is stated by Division Algorithm Theorem, for any integer $$a$$ and any positive integer $$b$$ there exists two unique integers such that $$a = bq + r$$, where $$a,b,q,r$$ are dividend, divisor, quotient and remainder respectively and $$r$$ is greater than or equal to $$0$$ and less than $$b$$.