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 $\mathrm{p}^{\alpha}$ and $(\mathrm{p}+2)^{\beta}$ divide $\Delta$, is ______.

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Determinants

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 ______.

Answer: 4

Factor out common terms from each row:

Δ = p! \cdot (p + 1)! \cdot (p + 2)! 1 p + 1 (p + 2) (p + 1) 1 p + 2 (p + 3) (p + 2) 1 p + 3 (p + 4) (p + 3)

Performing row operations (

R_{2} \to R_{2} - R_{1}

R_{3} \to R_{3} - (p + 1) R_{2}

) and simplifying, the determinant evaluates to 2.

Thus

Δ = 2 \cdot p! \cdot (p + 1)! \cdot (p + 2)!

.

Since

p

and

p + 2

are twin primes (e.g., 3 and 5),

p!

contains

p

once,

(p + 1)!

contains

p

once, and

(p + 2)!

contains

p

once, giving

α = 3

. Similarly,

(p + 2)!

contains

(p + 2)

once, so

β = 1

.

Therefore

α + β = 3 + 1 = 4

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Integer answer