请教prep模考软件上的一道数学题

kittycat23

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 100
B, between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. great than 40

为什么选E呢？谢谢大家了。

dontwannalose

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.
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以上是manhattang上的解释，不过我不是很明白，应为在1-50的prime no. 中说不定有一个数字（e.g. 3）在加1后刚好可以被除。。。（当然，我数学不怎么样，so long as u understand what it means...)
Since the smallest prime number that can be a factor of h(100) + 1 has to be greater than 50, The correct answer is E.

dontwannalose

just saw the further explaination:
the idea is this: if a number is divisible by some prime p, then the next multiple of p will be p units bigger. for instance, 75 is divisible by 5. this means that the next greatest multiple of 5 is 80, which is 5 units away.
hopefully, this fact is clear. once you realize this, it follows that consecutive integers can't share ANY primes, because they're only 1 unit apart (too close together to work for any common factor except 1, which is trivially a factor of any integer at all, anywhere).
that's the basis for saying h(100) + 1 has no prime factors below 50. if it did, then you'd have two multiples of the same prime 1 unit apart, and that's impossible.

kittycat23

It makes sense, but h(100)+1=(2^50) × 50=(2^50) × (2*5^2)=2^51*5^2, so the prime factor should be 5, right? 我不明白,

大荣

It makes sense, but h(100)+1=(2^50) × 50=(2^50) × (2*5^2)=2^51*5^2, so the prime factor should be 5, right? 我不明白,

-- by 会员 kittycat23 (2010/2/12 11:40:28)

LZ的式子写的不对，是h(100)有这个规律，不是加1，而且后面那个是50阶乘，不是50本身.
h(100)=(2^50) × 50!,  所以1-50都是h(100)的因子，那么1-50中的prime自然也是h(100)的prime factor。

翻译LS转的mahanttan解释：考虑1-50中的任一个prime，比如最小的prime 2，h(100)整除2，那么下一个整除2的是h(100)+2，所以h(100)+1肯定不能整除2. 其他小于50的prime类似。综合以上，所有小于50的prime都不可能是h(100)+1的prime factor。

kittycat23

谢谢大家的帮助！