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UID631941在线时间 小时注册时间2011-5-15最后登录1970-1-1主题帖子性别保密 
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| 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.
 
 the smallest prime number that can be a factor of h(100) + 1 has to be greater than 50,
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