HMMT 十一月 2025 · 冲刺赛 · 第 9 题

HMMT November 2025 — Guts Round — Problem 9

[7] Suppose S and T are two sets of distinct positive integers, each with 15 elements, such that S and T have no elements in common. Further suppose sum( S ) = sum( T ) = k, where sum( A ) denotes the sum of the elements of A . Compute the minimum possible value of k . © 2025 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2025, November 08, 2025 — GUTS ROUND Organization Team Team ID# 20

解析

Observe that 1800 + 225 = 2025, so n = 3600 works. Now assume for sake of contradiction that n < 3600. Since 2025 is odd, the two divisors that sum to 2025 must be distinct. Note that 2 ( n/ 3) + ( n/ 4) = 7 n/ 12, which is less than 2025 unless n = 59 , which clearly fails; thus, one of the divisors must be either n or n/ 2. If n is one of the divisors, let the other one be n/d for some positive integer d . Then, 2025 = n + n/d = n · ( d + 1) /d , so ( d + 1) /d = 2025 /n must be a quotient of two squares. But d + 1 and d cannot both be squares, contradiction. © 2025 HMMT If n/ 2 is one of the divisors, let the other one be n/d for some positive integer d . Then, n n d/ 2 + 1 d/ 2 + 1 2025 2025 = + = n = ⇒ = . 2 d d d n Since n is even, it is divisible by 4, so n/ 2 is even. We cannot have both n/ 2 and n/d even as they sum to 2025, so n/d is odd; hence, d is divisible by 4. It follows that d/ 2 + 1 and d are relatively prime (as the former is odd), so they are both squares. The above equation shows we want d to be as small as possible to minimize n ; while d = 4 fails, d = 16 works, which yields n = 3600 . This covers both cases. Hence, no smaller n exists. © 2025 HMMT