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