HMMT 二月 2023 · 团队赛 · 第 1 题

HMMT February 2023 — Team Round — Problem 1

[30] For any positive integer a , let τ ( a ) be the number of positive divisors of a . Find, with proof, the largest possible value of 4 τ ( n ) − n over all positive integers n .

解析

[30] For any positive integer a , let τ ( a ) be the number of positive divisors of a . Find, with proof, the largest possible value of 4 τ ( n ) − n over all positive integers n . Proposed by: Vidur Jasuja Answer: 12 n n Solution: Let d be the number of divisors of n less than or equal to . Then, τ ( n ) − 3 ≤ d ≤ = ⇒ 4 4 4 τ ( n ) − n ≤ 12 . We claim the answer is 12 . This is achieved by n = 12 . Remark. It turns out that n = 12 is the only equality case. One can see this by analyzing when exactly n τ ( n ) = + 3. 4