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 __________.

Solution

Maths sir · 5 steps

Pull a factorial out of each row, and the leftover determinant collapses to just 2.

1. Take out factorials. Take p! out of row 1, (p+1)! out of row 2 and (p+2)! out of row 3. Each row then becomes 1, (n+1), (n+1)(n+2), with n = p, p+1, p+2 in turn.

   Δ = 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)|

2. Subtract rows. Do R₃ → R₃ - R₂, then R₂ → R₂ - R₁. This gives R₂ = (0, 1, 2(p+2)) and R₃ = (0, 1, 2(p+3)). Expanding along the first column gives 2(p+3) - 2(p+2) = 2, so Δ = 2 p! (p+1)! (p+2)!.
3. Count powers of p. Since p and p+2 are both prime, p ≥ 3 and p+2 < 2p. So each of p!, (p+1)! and (p+2)! contains the factor p exactly once, and 2 is not a multiple of p. This gives α = 3.
4. Count powers of p+2. p+2 is a prime larger than every factor of p! and (p+1)!, so only (p+2)! contains it, and only once. This gives β = 1.
5. So. α + β = 3 + 1 = 4.

The idea to keep. When the rows of a determinant are consecutive factorials, take the smallest factorial out of each row first. The polynomial determinant that is left usually reduces to a constant.

Answer from the NTA final answer key. Solution by PrepWiser.