# $a$ and $b$ are two integers such that $10a+b=5$ and $p(x)=x^2+ax+b$.The value of n such that $p(10)p(11)=p(n)$

$\begin{array}{1 1}(A)\;110&(B)\;100\\(C)\;115&(D)\;105\end{array}$

Let $\alpha$ and $\beta$ be the roots of $p(x)=0$
$\Rightarrow p(x)=(x-\alpha)(x-\beta)$
$p(10)p(11)=(10-\alpha)(10-\beta)(11-\alpha)(11-\beta)$
$\Rightarrow (10-\alpha)(11-\beta)(10-\beta)(11-\alpha)$
$\Rightarrow [110-10(\alpha+\beta)-\alpha+\beta\alpha][110-10(\alpha+\beta)-\beta+\beta\alpha]$
$\Rightarrow (110+10a+b-\alpha)(110+10a+b-\beta)$
$\Rightarrow (115-\alpha)(115-\beta)$
$\Rightarrow p(115)$
Hence (C) is the correct answer.