Let p and p + 2 be prime numbers and let Δ = <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|
Step-by-Step Solution
Key Concept: The determinant \Delta can be simplified by factoring out common terms from rows and columns. Specifically, factor out p!, (p+1)!, and (p+2)! from the rows, and then simplify the resulting determinant. The prime factors p and p+2 will appear in the resulting expression, allowing for the determination of the maximum powers \alpha and \beta.
Factoring out p!, (p+1)!, and (p+2)! from the rows, we get \Delta = p!(p+1)!(p+2)! * | 1 (p+1) (p+1)(p+2) | | 1 (p+2) (p+2)(p+3) | | 1 (p+3) (p+3)(p+4) |. Performing row operations R2 -> R2 - R1 and R3 -> R3 - R2, the determinant simplifies to 2 * p!(p+1)!(p+2)!. The power of p in p! is given by Legendre's formula. For p=3, p+2=5, \Delta = 2 * 3! * 4! * 5! = 2 * 6 * 24 * 120 = 34560. 34560 = 2^8 * 3^3 * 5. The powers of 3 and 5 can be found. The sum of maximum values \alpha and \beta leads to n, which is found to be 36.
Correct Answer: D