Question

If $n^2-10n+21=p$ where $p$ is a prime number and $n\in N,$ then $n$ can be equal to

• A. $5$
• B. $4$
• C. $7$
• D. $8$

Answer 1

The correct option is D $8$

$n^2-10n+21=p$
$\Rightarrow n^2-7n-3n+21=p$
$\Rightarrow (n-3)(n-7)=p$
Since, one factor for $p$ will be definitely $1.$
So, $n-7=1(\because p>0)$
$\Rightarrow n=8$