JEE MAIN - Mathematics (2022 - 29th July Morning Shift - No. 21)
Let p and p + 2 be prime numbers and let
$$ \Delta=\left|\begin{array}{ccc} \mathrm{p} ! & (\mathrm{p}+1) ! & (\mathrm{p}+2) ! \\ (\mathrm{p}+1) ! & (\mathrm{p}+2) ! & (\mathrm{p}+3) ! \\ (\mathrm{p}+2) ! & (\mathrm{p}+3) ! & (\mathrm{p}+4) ! \end{array}\right| $$
Then the sum of the maximum values of $$\alpha$$ and $$\beta$$, such that $$\mathrm{p}^{\alpha}$$ and $$(\mathrm{p}+2)^{\beta}$$ divide $$\Delta$$, is __________.
Explanation
$$\Delta = \left| {\matrix{ {p!} & {(p + 1)!} & {(p + 2)!} \cr {(p + 1)!} & {(p + 2)!} & {(p + 3)!} \cr {(p + 2)!} & {(p + 3)!} & {(p + 4)!} \cr } } \right|$$
$$ = p!\,.\,(p + 1)!\,.\,(p + 2)!\left| {\matrix{ 1 & {(p + 1)} & {(p + 1)(p + 2)} \cr 1 & {(p + 2)} & {(p + 2)(p + 3)} \cr 1 & {(p + 3)} & {(p + 3)(p + 4)} \cr } } \right|$$
$$ = p!\,.\,(p + 1)!\,.\,(p + 2)!\left| {\matrix{ 1 & {p + 1} & {{p^2} + 3p + 2} \cr 0 & 1 & {2p + 4} \cr 0 & 1 & {2p + 6} \cr } } \right|$$
$$ = 2(p!)\,.\,\left( {(p + 1)!} \right)\,.\,\left( {(p + 2)!} \right)$$
$$ = 2(p + 1)\,.\,{(p!)^2}\,.\,\left( {(p + 2)!} \right)$$
$$ = 2{(p + 1)^2}\,.\,{(p!)^3}\,.\,\left( {(p + 2)!} \right)$$
$$\therefore$$ Maximum value of $$\alpha$$ is 3 and $$\beta$$ is 1.
$$\therefore$$ $$\alpha + \beta = 4$$
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