HMMT 二月 2003 · 冲刺赛 · 第 14 题
HMMT February 2003 — Guts Round — Problem 14
题目详情
- [7] A positive integer will be called “sparkly” if its smallest (positive) divisor, other than 1, equals the total number of divisors (including 1). How many of the numbers 2 , 3 , . . . , 2003 are sparkly?
解析
- A positive integer will be called “sparkly” if its smallest (positive) divisor, other than 1, equals the total number of divisors (including 1). How many of the numbers 2 , 3 , . . . , 2003 are sparkly? Solution: 3 Suppose n is sparkly; then its smallest divisor other than 1 is some prime p . Hence, e e e 1 2 r n has p divisors. However, if the full prime factorization of n is p p · · · p , the 1 2 r number of divisors is ( e + 1)( e + 1) · · · ( e + 1). For this to equal p , only one factor 1 2 r can be greater than 1, so n has only one prime divisor — namely p — and we get p − 1 p − 1 e = p − 1 ⇒ n = p . Conversely, any number of the form p is sparkly. There are 1 1 2 4 just three such numbers in the desired range (2 , 3 , 5 ), so the answer is 3.