官方模考中的两道题，求大神讲解

CaptainZHF

1.At a dinner party, 5 people are tp be seated around a circular table. Two seating arrangements are considered different only when the positions of the people are different relative to each other. What is the total member of different possible seating arrangements for the group.正确答案，24 。不知道怎么来的， 读了题并不很明白这个“when the positions of the people are different relative to each other”  

2.For every positive even integer n , h(n) is defined to be the product of all the even integers from 2 to n, if p is the smallest prime factor of h(100)+1 ,then p is?
答案是 greater than 40..   这类质数的题很头疼。

PS， 官方免费那两套模考和实战差不多吗？ 数学错了7个 语文错了14个 Q50 V38 730 ...........数学错7个还能50呢？？？

跳跃的癫痫

个人拙见 第一题那句话的意思是 几个人都向左移动一个位置 但是大家的相对位置不变时 这样是不能认同为与移动之前不一样的 所以只能大家的相对位置发生变化时才被认为是不同的情况。所以这个题可以以其中一个人为参考系 假设他不行 其他四个位置四个人随便坐 a44 一共24种
第二题你看看最近cd的贴我有看到大神给了很好的解答

跳跃的癫痫

假设他不动…手残

CaptainZHF

跳跃的癫痫 发表于 2017-5-6 10:54
个人拙见 第一题那句话的意思是 几个人都向左移动一个位置 但是大家的相对位置不变时 这样是不能认同为与移 ...

谢谢

Mandy66

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.
曼哈顿论坛上面找到的，因为 h(100) + 1 =2*(1*2*3*...*50)+1, 除以所有50以下的数都会有余数1,希望对楼主有点帮助

CaptainZHF

Mandy66 发表于 2017-5-6 16:33
Guest, this is definitely a difficult number properties question. Let's first consider the prime fac ...

明白了，谢谢