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

Hint: 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,
\[Q = {x^2} + x + 1\]
\[P = {x^{2007}} - x \\
\Rightarrow P = \left( {x - 1} \right)\left( {x + {x^2} + {x^3} + {x^4} + \ldots {x^{2006}}} \right)\]
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 \[b\].