- UID
- 503684
- 在线时间
- 小时
- 注册时间
- 2010-1-18
- 最后登录
- 1970-1-1
- 主题
- 帖子
- 性别
- 保密
|
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.
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.
Hope that helps -Dan |
|