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HMMT 二月 2024 · 冲刺赛 · 第 29 题

HMMT February 2024 — Guts Round — Problem 29

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [16] For each prime p , a polynomial P ( x ) with rational coefficients is called p -good if and only if there p exist three integers a , b , and c such that 0 ≤ a < b < c < and p divides all the numerators of P ( a ) , 3 P ( b ) , and P ( c ) , when written in simplest form. Compute the number of ordered pairs ( r, s ) of rational 3 2 numbers such that the polynomial x + 10 x + rx + s is p -good for infinitely many primes p .
解析
  1. [16] For each prime p , a polynomial P ( x ) with rational coefficients is called p -good if and only if there p exist three integers a , b , and c such that 0 ≤ a < b < c < and p divides all the numerators of P ( a ) , 3 P ( b ) , and P ( c ) , when written in simplest form. Compute the number of ordered pairs ( r, s ) of rational 3 2 numbers such that the polynomial x + 10 x + rx + s is p -good for infinitely many primes p . Proposed by: Pitchayut Saengrungkongka Answer: 12 Solution: By Vieta, the sum of the roots is − 10 (mod p ) . However, since the three roots are less than ′ ′ ′ p/ 3 , it follows that the roots are ( p − a ) / 3 , ( p − b ) / 3 , ( p − c ) / 3 , where there are finitely many choices ′ ′ ′ a < b < c . By pigeonhole, one choice, say ( u, v, w ) must occur for infinitely many p . We then get that the roots of P are − u/ 3 , − v/ 3 , and − w/ 3 . Moreover, we must have that u, v, w are all 1 (mod 3) or all 2 (mod 3) , and by Vieta, we have u + v + w = 30 . The polynomial is then uniquely determined by u, v, w . Thus, it suffices to count triples u < v < w of positive integers such that u , v , w are all 1 (mod 3) or all 2 (mod 3) and that u + v + w = 30 . It’s not very hard to list them all now. When u, v, w ≡ 1 (mod 3) , there are 7 triples: (1 , 4 , 25) , (1 , 7 , 22) , (1 , 10 , 19) , (1 , 13 , 16) , (4 , 7 , 19) , (4 , 10 , 16) , and (7 , 10 , 13) . When u, v, w ≡ 2 (mod 3) , there are 5 triples: (2 , 5 , 23) , (2 , 8 , 20) , (2 , 11 , 17) , (5 , 8 , 17) , and (5 , 11 , 14) . Hence, the answer is 7 + 5 = 12 .