Question

Suppose $p\left(x\right)$ is a polynomial with integer coefficients. The remainder when $p\left(x\right)$ is divided by $x-1$ is $1$ and the remainder when $p\left(x\right)$ is divided by $x-4$ is $10$. If $r\left(x\right)$ is the remainder when $p\left(x\right)$ is divided by $(x-1)(x-4)$, then the value of $r\left(2006\right)$, is

• A. $6018$
• B. $6016$
• C. $6020$
• D. None of these

Answer 1

Answer: (B)

$6016$

Let $Q_1,Q_2$ be the quotients when $p\left(x\right)$ is divided by $(x-1),(x-4)$ respectively.
$p\left(x\right)=(x-1)Q_1\left(x\right)+1$ ....(1)
$p\left(x\right)=(x-4)Q_2\left(x\right)+10$ .....(2)
$\Rightarrow (x-4)p\left(x\right)=(x-4)(x-1)Q_1\left(x\right)+(x-4)$ ....(3)
and $(x-1)p\left(x\right)=(x-1)(x-4)Q_2\left(x\right)+10x-10$ ....(4)
Subtracting (3) from (4), we get
$\Rightarrow p\left(x\right)=(x-1)(x-4)\frac{\left[Q_2\right(x)-Q_1(x\left)\right]}{3}+\frac{9x-6}{3}$
Here, $p\left(x\right)$ when divided by $(x-1)(x-4)$, remainder $=\frac{9x-6}{3}$
$\Rightarrow r\left(x\right)=\frac{9x-6}{3}=3x-2$
$\Rightarrow r\left(2006\right)=3\times 2006-2=6016$