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标题: 请教1道GMAT PREP的题目 [打印本页]

作者: cathy_ly 时间: 2007-5-16 10:37
标题: 请教1道GMAT PREP的题目

1。 For every positive even interger 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. beteween 2 and 10

b.between 10 and 20

c.between 20 and 30

d.between 30 and 40

e.greater than 40

answer: e

这种题目的解题思路是什么呀？

作者: cuckoo121 时间: 2007-5-16 11:24
i would like to know too....someone help please..

作者: specky 时间: 2007-5-16 12:18

for every positive even integer n, the fonction h(n)
is defined to be the product of all the even integers from 2 to n,
inclusive.

So we could write the following:
h(2)=2=2*1
h(4)=2*4=(2*1)*(2*2)=2^2*(1*2)
h(6)=2*4*6=(2*1)*(2*2)*(2*3)=2^3*(1*2*3)
h(8)=2*4*6*8=(2*1)*(2*2)*(2*3)*(2*4)=2^4*(1*2*3*4)
...
You can see if n=2m then
h(n)=2^m*(1*2*...*m)=2^m*m!

This is exactly what you need to know to solve the problem.
At a minimum the smallest prime factor is 53.

For any factorial + 1 the smallest factor (apart from 1) is greater then any of the members of the factorial

2! + 1 = 3: smallest factor is 3
3! + 1 = 7: smallest factor is 7
4! + 1 = 25: smallest factor is 5
5! + 1 = 121: smallest factor is 11

Therefore, the solution has to be >50 (choose e)

作者: cathy_ly 时间: 2007-5-17 08:00

以下是引用specky在2007-5-16 12:18:00的发言：

This is exactly what you need to know to solve the problem.
At a minimum the smallest prime factor is 53. ?? 为何是53？

For any factorial + 1 the smallest factor (apart from 1) is greater then any of the members of the factorial

2! + 1 = 3: smallest factor is 3
3! + 1 = 7: smallest factor is 7
4! + 1 = 25: smallest factor is 5
5! + 1 = 121: smallest factor is 11

Therefore, the solution has to be >50 (choose e)

谢谢回复，但是红色那个地方还是不明白

作者: biggeo 时间: 2007-5-17 15:14
同问.
难道是归纳法得出得?如何证明啊?

作者: specky 时间: 2007-5-17 16:35
h(n)=2^m*(1*2*...*m)=2^m*m!
from above, since m is 50!, therefore anything between 1 to 50 is its factor, min prime factor has to be greater than 50.

作者: biggeo 时间: 2007-5-17 23:38
Hi specky,
You're right on
h(n)=2^m*m!, and it's true that 1 to 50 is h(n)'s factor
But how can you determine to min prime factor of h(n)+1??

作者: gonghao 时间: 2007-5-17 23:58

偶数都是2k，k属于自然数

连续偶数的乘积，等于(2^k)*K!

因为2k=100则应该有50项偶数

则应该是K=50即，h（100）=50!*2^50=50!*(4^5)^5

h（100）+1与h（100）互质

则h（100）+1没有h（100）的任何因子，除了1

1到50里的数字一个也不能出现，否则2者不能互质

所以只能去大于50的最小质数（53，我想楼上的53就是这样得来的，注意不是51=3*17其不是质数）

1在选项里排除了

只能选大于50的数字，

E凑合

[此贴子已经被作者于2007-5-18 0:04:23编辑过]

作者: mbamom 时间: 2007-5-18 00:15
thanks for the explanation, guess if I were to solve this problem in the exam, I would skip it

作者: biggeo 时间: 2007-5-18 00:18
偶像!

"h（100）+1与h（100）互质" 是关键啊

作者: cathy_ly 时间: 2007-5-18 05:29

明白了，谢谢指点。崇拜！

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