HMMT 十一月 2008 · 团队赛 · 第 3 题

英文解析

3. Let p be a prime. Given a sequence of positive integers b through b , exactly one of which
   1 nis divisible by p , show that when
   1 1 1 + + . . . +
   b b b
   1 2 nis written as a fraction in lowest terms, then its denominator is divisible by p . Use this factto 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 ordern nof addition doesn’t matter. We can then write
   1 1 1 a + + . . . + =
   b b b b
   1 2 n − 1
   kpa + ba 1
   where b is not divisible by p (since none of the b are). But then + = . Since bib kp kpbis not divisible by p , kpa + b is not divisible by p , so we cannot remove the factor of p from the denominator. In particular, p cannot be juicy as 1 can be written as , which has a 1
   1 11
   denominator not divisible by p , whereas being juicy means we have a sum + . . . + = 1,
   b b
   1 nwhere b < b < . . . < b = p , and so in particular none of the b with i < n are divisible by
   1 2 n ip .