HMMT 二月 2019 · 冲刺赛 · 第 23 题

HMMT February 2019 — Guts Round — Problem 23

[ 12 ] Find the smallest positive integer n such that 2 ··· 2 2 2 > (( · · · ((100!)!)! · · · )!)! . ︸︷︷︸︸︷︷︸ n 2’s 100 factorials

解析

[ 12 ] Find the smallest positive integer n such that 2 ··· 2 2 2 > (( · · · ((100!)!)! · · · )!)! . ︸︷︷︸︸︷︷︸ n 2’s 100 factorials Proposed by: Zack Chroman Answer: 104 2 2 2 2 2 a 2 Note that 2 > 100 . We claim that a > b = ⇒ 2 > ( b !) , for b > 2. This is because a 2 b 2 > b ⇐⇒ a > 2 b log ( b ) , 2 2 b and log ( b ) < b / 2 for b > 2. Then since b > b ! this bound works. Then 2 ... 2 2 2 2 (2 ) > ((((100!)!)!)! . . . ) ︸︷︷︸︸︷︷︸ m 2’s m − 4 factorials n for all m ≥ 4 by induction. So n = 104 works. The lower bound follows from the fact that n ! > 2 for 2 2 n > 3 , and since 100 > 2 , we have 100 2 2 ... ... 2 2 (((100!)!)!)! . . . ) > 2 > 2 . ︸︷︷︸︸︷︷︸︸︷︷︸ 100 2’s 103 2’s 100 factorials