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← JEE Main 2022 · 29 Jul · Shift 1
MathematicsInteger+4 / -1Question 27 of 90

JEE Main 2022 · 29 Jul · Shift 1, Mathematics question 27

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=\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 α and β, such that pαp^\alpha and (p+2)β(p+2)^\beta divide Δ, is _______.

Answer and worked solution

Official answer: 4

Factor rows. Taking p!p!, (p+1)!(p+1)!, (p+2)!(p+2)! out of the rows leaves rows 1, p+1+j, (p+1+j)(p+2+j)1,\ p+1+j,\ (p+1+j)(p+2+j) in shape; subtracting rows gives determinant 2. So Δ=2 p! (p+1)! (p+2)!\Delta=2\,p!\,(p+1)!\,(p+2)!.

Powers. For an odd prime p (and p+2<2pp+2<2p), p divides each factorial once: α=3\alpha=3. p+2p+2 appears only in (p+2)!(p+2)!: β=1\beta=1. Sum 4.

Answer: 4