HMMT 二月 2022 · 冲刺赛 · 第 8 题

HMMT February 2022 — Guts Round — Problem 8

[6] For any positive integer n , let τ ( n ) denote the number of positive divisors of n . If n is a positive 2 7 τ ( n ) τ ( n ) integer such that = 3, compute . τ ( n ) τ ( n ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2022, February 19, 2022 — GUTS ROUND Organization Team Team ID#

解析

[6] For any positive integer n , let τ ( n ) denote the number of positive divisors of n . If n is a positive 2 7 τ ( n ) τ ( n ) integer such that = 3, compute . τ ( n ) τ ( n ) Proposed by: Sean Li Answer: 29 e e e 1 2 k Solution: Let the prime factorization of n be n = p p · · · p . Then, the problem condition is 1 2 k equivalent to k Y 2 e + 1 i = 3 . e + 1 i i =1 2 x +1 3 Note that since ≥ 1 . 5 for x ≥ 1, and 1 . 5 > 3 , we have k ≤ 2. Also, k = 1 implies 2 e + 1 = 1 x +1 3( e + 1) , which implies e is negative. Thus, we must have k = 2. Then, our equation becomes 1 1 (2 e + 1)(2 e + 1) = 3( e + 1)( e + 1) , 1 2 1 2 2 4 which simplifies to ( e − 1)( e − 1) = 3. This gives us e = 2 and e = 4. Thus, we have n = p q for 1 2 1 2 7 14 28 τ ( n ) τ ( p q ) 15 · 29 primes p and q , so = = = 29. 2 4 τ ( n ) τ ( p q ) 3 · 5