标题: 求各位大神一道数学题~ [打印本页]

---

作者: xuchaodream    时间: 2012-3-15 07:13     标题: 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

---

作者: xuchaodream    时间: 2012-3-15 07:14

补充下，答案是E

---

作者: voleroiseau    时间: 2012-3-15 20:14

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.

---

作者: 天叉包子    时间: 2012-3-16 06:47

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

---

作者: voleroiseau    时间: 2012-3-16 20:27

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

---

作者: supercury    时间: 2012-3-22 20:34

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

---

欢迎光临 国际顶尖MBA申请交流平台--TOPWAY MBA (http://forum.topway.org/)

Powered by Discuz! 7.2