HMMT 十一月 2018 · 冲刺赛 · 第 13 题

13. [ 9 ] Find the smallest positive integer n for which 2 1!2! · · · ( n 1)! > n ! . Proposed by: Andrew Gu Answer: 8 2 Dividing both sides by n ! , we obtain 1!2! ... ( n 3)!( n 2)!( n 1)!

1 [ n ( n 1)!] [ n ( n 1)( n 2)!] 1!2! ... ( n 3)! 1 2 n ( n 1) 2 1!2! ... ( n 3)! > n ( n 1) Factorials are small at first, so we can rule out some small cases: when n = 6, the left hand side is 2 1!2!3! = 12, which is much smaller than 6 · 5. (Similar calculations show that n = 1 through n = 5 do not work. either.) 2 Setting n = 7, the left-hand side is 288, which is still smaller than 7 · 6. However, n = 8 gives 34560 > 448, so 8 is the smallest integer for which the inequality holds.