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求各位大神一道数学题~

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
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补充下,答案是E

TOP

h(100)=2*4*6*...*100,
两个相邻的数不会有共同的质因数,
h(100)的最大质因数为47,并且包含了所有小于47的质数。
而h(100)+1使得原来的这些质因数都不能整除,所以要找质因数要一定要大于47.

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h(100)的最大质因数为47,并且包含了所有小于47的质数
为啥呀

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因为h(100)=2*4*...*100,其中包括了所以小于50的数的倍数,所以每个小于50的数都是h(100)的因数,
47是小于50的最大的质数
所以,h(100)的最大质因数是47
我觉得是这样

TOP

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

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