HMMT 二月 2022 · ALGNT 赛 · 第 9 题

HMMT February 2022 — ALGNT Round — Problem 9

Suppose P ( x ) is a monic polynomial of degree 2023 such that 1 2023 P ( k ) = k P 1 − k a for every positive integer 1 ≤ k ≤ 2023. Then P ( − 1) = , where a and b relatively prime integers. b Compute the unique integer 0 ≤ n < 2027 such that bn − a is divisible by the prime 2027.

解析

Suppose P ( x ) is a monic polynomial of degree 2023 such that 1 2023 P ( k ) = k P 1 − k a for every positive integer 1 ≤ k ≤ 2023. Then P ( − 1) = , where a and b relatively prime integers. b Compute the unique integer 0 ≤ n < 2027 such that bn − a is divisible by the prime 2027. Proposed by: Akash Das Answer: 406 n n − 1 Solution: Let n = 2023. If P ( x ) = x + a x + · · · + a , then let n − 1 0 1 n n n n R ( x ) = x P (1 − ) = ( x − 1) + a ( x − 1) x + · · · + a x . n − 1 0 x Then, note that Q ( x ) = P ( x ) − R ( x ) is a polynomial of degree at most n , and it has roots 1 , 2 , . . . , n , so we have Q ( x ) = k ( x − 1) · · · ( x − n ) for some real constant k . Now we determine P ( x ) in terms of 1 Q ( x ). If g ( x ) = 1 − 1 /x , then g ( g ( x )) = and g ( g ( g ( x ))) = x . Therefore, we have 1 − x 1 n P ( x ) − x P 1 − = Q ( x ) x n 1 1 1 1 P 1 − − 1 − P = Q 1 − x x 1 − x x n 1 1 1 P − P ( x ) = Q . 1 − x 1 − x 1 − x n n Adding the first equation, x times the second, and ( x − 1) times the third yields x − 1 1 n n 2 P ( x ) = Q ( x ) + x Q + ( x − 1) Q , x 1 − x so k P ( x ) = ( x − 1)( x − 2) · · · ( x − n ) + (0 x − 1)( − 1 x − 1) · · · ( − ( n − 1) x − 1) 2

( − 1 x + 0)( − 2 x + 1) · · · ( − nx + ( n − 1)) . Therefore, k P ( − 1) = ( − ( n + 1)! + 0 + (2 n + 1)!!) . 2 Also, since P is monic, we know that k 1 = (1 + 0 − n !) , 2 so (2 n − 1)!! − ( n + 1)! P ( − 1) = . 1 − n ! Modulo 2027, ( n + 1)! = 2026! / (2026 · 2025) ≡ − 1 / ( − 1 · − 2) ≡ − 1 / 2 and n ! = ( n + 1)! / 2024 ≡ 1 / 6. Also, (2 n + 1)!! ≡ 0. So our answer is 1 / 2 3 2030 = ≡ = 406 . 1 − 1 / 6 5 5