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  4. Let p and p +2 be prime numbers and let Δ=| p ! (p+1) ! (p+2) ! (p+1) ! (p+2) ! (p+3) ! (p+2) ! (p+3) ! (p+4) ! | Then the sum of the maximum values of α and β, such that p α and ( p +2)β divide Δ, is

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

JEE MainJEE Main 2022Determinants

Solution:

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