[求助]一道数学题

For every positive even integer n, the function h(n) is defined to be 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

(A) between 2 and 10

(B) between 10 and 20

(C) between 20 and 30

(D) between 30 and 40

(E) greater than 40

以前见过的，但是忘了算法了，请教，谢谢。

I think the answer should be E

From 2-40 the bigest prime is 37, 37X2=74, 74 is the factor of h(100)(74 is one even integer between 2-100), so 37 must be the factor of H(100), then 37 is not the factor of H(100)+1(use the same way we can know that other prime integers which are smaller than 37 must be the factor of h(100), so not the factor of h(100)+1).  so, the smallest prime integer factor should be greater than 40

来个简单的方法，嘿嘿

h(n)是偶数序列2,4,6,。。。，100的乘积，都提一个2出来

h(n)=2^50*50!

h(n)+1=2^50*50!+1

即h(n)+1有质因子的话，必大约50。条件加强，也就大于40了

谢谢楼上两位。

这里的推理根据是，

如果某数是h(n)的因子，那么这个数不会是h(n)+1 的因子，

就是这点我没想明白。

＋1之后大概是个什么情况，没想清楚。