标题: 请教Prep2006上的一道数学 [打印本页] 作者: sunnygirl_jj 时间: 2009-11-2 04:29 标题: 请教Prep2006上的一道数学 For every positive even integer n, h(n) is defined as the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is
Answer: greater than 40
这道题用什么方法解？作者: 切尔西 时间: 2009-11-2 12:17
Guest, this is definitely a difficult number properties question. Let's first consider the prime factors of h(100). According to the given function, h(100) = 2*4*6*8*...*100
By factoring a 2 from each term of our function, h(100) can be rewritten as 2^50*(1*2*3*...*50).
"Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).
Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1. "
So because h(100) can be factored by every integer up to 50 means that h(100)+1 can't have a prime factor below 50