HMMT 二月 2019 · 冲刺赛 · 第 5 题

HMMT February 2019 — Guts Round — Problem 5

[ 4 ] Call a positive integer n weird if n does not divide ( n − 2)!. Determine the number of weird numbers between 2 and 100 inclusive.

解析

[ 4 ] Call a positive integer n weird if n does not divide ( n − 2)!. Determine the number of weird numbers between 2 and 100 inclusive. Proposed by: Yuan Yao Answer: 26 We claim that all the weird numbers are all the prime numbers and 4. Since no numbers between 1 and p − 2 divide prime p , ( p − 2)! will not be divisible by p . We also have 2! = 2 not being a multiple of 4. Now we show that all other numbers are not weird. If n = pq where p 6 = q and p, q ≥ 2, then since p and q both appear in 1 , 2 , . . . , n − 2 and are distinct, we have pq | ( n − 2)!. This leaves the only 2 2 case of n = p for prime p ≥ 3. In this case, we can note that p and 2 p are both less than p − 2, so 2 2 p | ( n − 2)! and we are similarly done. Since there are 25 prime numbers not exceeding 100, there are 25 + 1 = 26 weird numbers.