HMMT 十一月 2008 · 团队赛 · 第 3 题

HMMT November 2008 — Team Round — Problem 3

Let p be a prime. Given a sequence of positive integers b through b , exactly one of which 1 n is divisible by p , show that when 1 1 1

- . . . + b b b 1 2 n is written as a fraction in lowest terms, then its denominator is divisible by p . Use this fact to explain why no prime p is ever juicy.

解析

Let p be a prime. Given a sequence of positive integers b through b , exactly one of which 1 n is divisible by p , show that when 1 1 1

- . . . + b b b 1 2 n is written as a fraction in lowest terms, then its denominator is divisible by p . Use this fact to explain why no prime p is ever juicy. Solution: We can assume that b is the term divisible by p (i.e. b = kp ) since the order n n of addition doesn’t matter. We can then write 1 1 1 a
- . . . + = b b b b 1 2 n − 1 kpa + b a 1 where b is not divisible by p (since none of the b are). But then + = . Since b i b kp kpb is not divisible by p , kpa + b is not divisible by p , so we cannot remove the factor of p from 1 the denominator. In particular, p cannot be juicy as 1 can be written as , which has a 1 1 1 denominator not divisible by p , whereas being juicy means we have a sum + . . . + = 1, b b 1 n where b < b < . . . < b = p , and so in particular none of the b with i < n are divisible by 1 2 n i p .