HMMT 二月 2026 · 团队赛 · 第 10 题
HMMT February 2026 — Team Round — Problem 10
题目详情
- [55] Prove that there exists a real constant M such that for every prime p ≥ M and any positive 0 . 99 integer 2 ≤ m ≤ p − 1 , there exist positive integers a and b such that m ≤ a ≤ 1 . 01 m , p ≤ b ≤ p , and p divides ab − 1 . ©2026 HMMT
解析
- [55] Prove that there exists a real constant M such that for every prime p ≥ M and any positive 0 . 99 integer 2 ≤ m ≤ p − 1 , there exist positive integers a and b such that m ≤ a ≤ 1 . 01 m , p ≤ b ≤ p , and p divides ab − 1 . Proposed by: Pitchayut Saengrungkongka Answer: − 1 Solution: Let f ( k ) = k mod p , so p | kf ( k ) − 1 . p +1 First, we note that since p | mf ( m ) − 1 , we get that for all m ≥ 2 , mf ( m ) ≥ p + 1 , so f ( m ) ≥ . In m 0 . 01 0 . 01 particular, if m < p , then a = m and b = f ( m ) works. Henceforth, we assume that m > p . In particular, m > 1000 by taking M to be large enough. af ( a ) − 1 0 . 99 Assume for contradiction that f ( a ) ≤ p for all a ∈ [ m, 1 . 01 m ] . Consider the quotients as a p ranges across the interval [ m, 1 . 01 m ] . There are 0 . 01 m − 1 > 0 . 005 m numbers, each has size at most 0 . 99 af ( a ) − 1 1 . 01 m · p 2 m ≤ < . 0 . 01 p p p af ( a ) − 1 0 . 01 Therefore, there exists k such that = k for at least 0 . 001 p values a ∈ [ m, 1 . 01 m ] . Such a p 0 . 01 2 must divide pk + 1 , so we deduce that pk + 1 has more than 0 . 001 p divisors. Since pk + 1 < p , the following lemma gives a contradiction (by taking ε < 0 . 01 ). Lemma 1. For any real number ε > 0 , there exists a constant M (possibly depending on ε ) such that ε for all n > M , the number of divisor of n is at most n . e e e 1 2 k Proof. Enumerate all prime numbers as p < p < . . . . Let n = p p . . . p , so the number of divisors 1 2 1 2 k is ( e + 1)( e + 1) . . . ( e + 1) . We observe that 1 2 k 0 . 5 εe 1 /ε e i i • If p > 10 , then e + 1 ≤ 10 < p . i i i 0 . 5 εe i • Otherwise, there exists a constant C such that e + 1 < C p for all e ≥ 0 . i i i i i 1 /ε Suppose that m is the largest integer such that p ≤ 10 . Then we have that the number of divisors m of n is 0 . 5 εe εe 0 . 5 εe 0 . 5 ε 1 2 k ( e + 1)( e + 1) . . . ( e + 1) ≤ C C . . . C p p . . . p = C C . . . C n , 1 2 k 1 2 m 1 2 m 1 2 k 2 /ε so picking M > ( C C . . . C ) concludes. 1 2 m ©2026 HMMT