HMMT 十一月 2024 · 冲刺赛 · 第 24 题

HMMT November 2024 — Guts Round — Problem 24

[12] Let f ( x ) = x + 6 x + 6 . Compute the greatest real number x such that f ( f ( f ( f ( f ( f ( x )))))) = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT November 2024, November 09, 2024 — GUTS ROUND Organization Team Team ID#

解析

[12] Let f ( x ) = x + 6 x + 6. Compute the greatest real number x such that f ( f ( f ( f ( f ( f ( x )))))) = 0. Proposed by: Arul Kolla √ 64 Answer: 3 − 3 2 Solution: Observe that f ( x ) = ( x + 3) − 3. Now, we claim that k k 2 Claim 1. f ( x ) = ( x + 3) − 3 for all positive integers k . Proof. We use induction. The base case k = 1 is clear. To show the inductive step, note that k k 2 f ( x ) = ( x + 3) − 3 implies k k k +1 2 k +1 k 2 2 2 f ( x ) = f ( f ( x )) = f (( x + 3) − 3) = (( x + 3) − 3) + 3 − 3 = ( x + 3) + 3 . √ √ 64 64 6 64 Thus, if r is a real root of f , then ( r + 3) = 3, so r + 3 = ± 3, and hence r = ± 3 − 3. The √ 64 largest value of r is thus 3 − 3 .