新模考数学题-求助

期待蔚蓝

以下是引用julia_ggw在2005-12-27 4:28:00的发言：

这个例子比较清楚，那么本题就应该是C(4,1)C(3,1)C(2,1)C(1,1)=24

At a dinner party, 5 people are to be seated around a circular table, two seating arrangments are considered differently only when the positions of the people are different relative to each other. what is the total number of different possible seating arrangments for the group.

a) 5

b)10

c)24

d) 32

e) 120

解答的思路是什么？

期待蔚蓝

明白啦！

5个人，全排列是P55=5！

那么5个人的圆桌排列就是5！/5=4！

[此贴子已经被作者于2007-1-2 0:04:51编辑过]

期待蔚蓝

以下是引用期待蔚蓝在2006-12-26 15:51:00的发言：

谁能再给解释一下这道题目啊？

Last month 15 homes were sold in town x, the average sale price of the homes was $150000 and the median sale price was $130000, which of the following statement must be true.

1) at least one of the homes was sold for more than $165000

2) at least one of the homes was sold for more than $130000 and less than $150000

3) at least one of the homes was sold for less than $130000.

答 （1）

唉，自己解释吧，刚问过别人

一看到average,先把总数算出来，我们把后面一大串0先去掉，假设average15,median13，好吧
那么total=15(homes)*15(dollar)=225

又因为median=13，也就是说第8项=13

那么从小到大排的话前7个都不能大于第8项，也就是最大也只能是13，那么前8个的总合最大就是8*13=104，那么后7个最小的总合就是225-104=121

那么，后7个的均值就是121/7=17.28...

这样的话即使你后面7个一样的价，也都大于16.5，即使有一个比均值小甚至小于16.5，那么肯定对应的，至少会有一个比均值还大，所以至少会有一个大于16.5，也就是条件1满足

再看条件2

如果前8个都是13 后面7个总合是121，价钱都一样的话就是每个17多，完全可以没有在13-15之间的
再看3他没说要求15个房子价钱都不一样

kaochieh

thanks!

hhdty

顶

goo_puu3

Circular table's Permutation can be seen as N-1 的队列 Permutation.

yuching730

ding

bwainecho

1.  Because h(100) = 2 * (2*2) * (2*3) * (2*4) .... (2*47) (2*48) (2*49) (2*50)

47 is the maximal prime factor of h(100).  All other prime numbers < 47 are also the factors of h(100).
So all prime numbers < 47 can NOT be factors of h(100)+1.

=> P has to be > 47


2.   think this situation that 14 houses are sold at 130 000, the last house is very very expensive to make average to 150000, so both condition 2 and 3 are not necessarily true

3.  P(5, 5) / 5;
    because
    ABCDE = BCDEA = CDEAB = DEABC = EABCD in a round table.

eileenmu木

up

hippo3000

以下是引用maple_leaf在2005-12-26 18:05:00的发言：

大家看看这道题怎么做?完全没有思路的说

For every positive even integer n, the function h(n) is defined to be the product of all the even integers form 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

一点思路都没有，盼思路讲解。

这题不要被题目吓倒，其实非常简单，思路是想h(100) 能被谁整除，由于H100是2-100偶数相乘，所以一定是50以下质数的倍数，因为任何50以下质数2倍都在2－100偶数里面。

如果h(100) 可以整除所有50以下质数，则h(100)+1一定不能被50以下质数整除，因为对于50以下的任何质数P你都可以这样改写 h(100)+1=P*K+1, k为正整数，因为P*K可以被P整除，所以P*K+1被P除，一定余1，故不可能整除。

故E为正确答案