请教Prep2006上的一道数学

sunnygirl_jj · 发表于 2009-11-2 04:29:21

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？

切尔西 · 发表于 2009-11-2 12:17:06

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

在google上找的，自己算了一下应该是正确的，遇到这个题好像没有什么简单的算法了，一开始也想了好久，不够没有办法，只好GOOGLE了
希望可以帮到你

sunnygirl_jj · 发表于 2009-11-3 02:45:48

真是好人啊，还帮我google了，太感谢了！
这道题我昨天晚上想通了，就是这个解法。哎，很多年没碰数论了，连解法都忘了。

请教Prep2006上的一道数学

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