Question

Let $p$ and $p+2$ be prime numbers and let
$\Delta =\left|\begin{array}{ccc}p!& (p+1)!& (p+2)!\\ (p+1)!& (p+2)!& (p+3)!\\ (p+2)!& (p+3)!& (p+4)!\end{array}\right|$
Then the sum of the maximum values of $\alpha$ and $\beta$, such that $p^{\alpha }$ and $(p+2{)}^{\beta }$ divide $\Delta$, is

Answer 1

Answer: 4

$\Delta =\left|\begin{array}{ccc}p!& (p+1)!& (p+2)!\\ (p+1)!& (p+2)!& (p+3)!\\ (p+2)!& (p+3)!& (p+4)!\end{array}\right|$
$\Delta =P!(P+1)!(P+2)!\left|\begin{array}{ccc}1& 1& 1\\ P+1& P+2& P+3\\ (P+2)(P+1)& (P+3)(P+2)& (P+4)(P+3)\end{array}\right|$
$\Delta =2P!(P+1)!(P+2)!$
Which is divisible by $P^{\alpha }\&(P+2{)}^{\beta }$
$\therefore \alpha =3,\beta =1$