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Matrices and Determinants question

2022 · 29 Jul · Shift 1 · Q45
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Matrices and Determinants question

2022 · 29 Jul · Shift 1 · Q45

JEE MainMathematicsMatrices and DeterminantsNumerical+4 / −1
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)!∣\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|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\alphaα and β\betaβ, such that pα\mathrm{p}^{\alpha}pα and (p+2)β(\mathrm{p}+2)^{\beta}(p+2)β divide Δ\DeltaΔ, is ‾\underline{\hspace{2cm}}​.
Numerical answer
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