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Prep模考里的题目请教

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楼主
发表于 2011-4-2 12:38:39 | 只看该作者 回帖奖励 |倒序浏览 |阅读模式
1. For every positive integer n, the function h(n) is defined to be the product of alll the even integers from 2 to n, inclusive. If p is the smallest prim factor of h(100)+1, the p is       ?
答案:more than 40

2. 五个人被安排在一个圆桌上就座,问有多少种分法?只要旁边的人相同都只算一种。
答案:24

第一题我完全没思路,所以不能说自己的想法。。。
第二题我开始选120,算法是5*4*3*2*1,后来看到错了以后想也许应该除以5,得到24. 但是想不明白为什么120里面包含同样的分配方法绕着桌子转一圈。期待解释!
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沙发
发表于 2011-4-2 12:55:29 | 只看该作者
h(100) = 2*4*6*8*. . .*94*96*98*100 = 2^50 *(1*2*3*4* . . . *47*48*49*50)

Apparently, h(100) have all the prime factors between 2 and 47.

h(100) and h(100) + 1 are two consecutive integers. As such, these two numbers are co-prime, meaning they do not share prime factors.

Therefore, h(100) + 1 do not have prime factors between 2 and 47. The smallest prime factor it could have is 53.
板凳
发表于 2011-4-2 12:58:34 | 只看该作者
Suppose you have ABCDE sitting on a table. Then you have "another" BCDEA sitting on the table. If you walk around the table, you 'will find these two arrangements are the same! So is CDEAB, DEABC, and EABCD. That's why you need to devide what you get from a linear arrangement by 5 in the circular case.
地板
 楼主| 发表于 2011-4-2 13:04:40 | 只看该作者
非常感谢sdcar这么快回复。
第一题很清楚,我看的很明白。
可是第二题我还是不太清楚为什么当我考虑从5个人中挑出1个坐在某个位置上,然后再从剩余4个里挑出一个,一直到所有人都安排到座位包含了绕桌子一圈。我明白如果包含,那应该除掉,因为算相同安排方法。
5#
发表于 2011-4-2 13:37:42 | 只看该作者
Linear arrangements of ABCDE, BCDEA, CDEAB, DEABC, and EABCD are all unique arrangements, in part because the head and toe positions are occupied by different letters.

Circular arrangements of ABCDE, BCDEA, CDEAB, DEABC, and EABCD are the same setup because the chairs around the table are not labeled (只要旁边的人相同都只算一种). In all five instances, A is always between B and E, B is always between A can C , etc.

The difference between circular and linear setups is the at the former has no head or tail positions.
6#
 楼主| 发表于 2011-4-2 22:34:09 | 只看该作者
我的问题没有表述清楚。但是我明白了你说的ABCDE和BCDEA等等是一种坐法。我现在想不明白的是为什么120的那种算法计算了所有这些重复的坐法,所以要除以5?
7#
发表于 2011-4-2 22:40:44 | 只看该作者
Because you assume the first one chose is the "head" and the last one you chose is the "tail"  around the table. In reality, neither head nor tail exist around the table.

To assign 5 people around a table, you do the following:
1) Pick one person and let it sits on one seat on the table.
2) Pick another from the remaining 4 and let it sit to the left of the first one: 4 possibilities
3) Pick another from the remaining 3; left; of the 2nd: 3 possibilities
4) Pick another, 2; left; 3rd: 2 possibilities.

Total = 4!
8#
 楼主| 发表于 2011-4-2 23:09:06 | 只看该作者
那选第一个人就坐的时候不是有5种可能性吗?
9#
发表于 2011-4-3 01:05:49 | 只看该作者
No. The first one we picked does not count. Think about in this way: Let's suppose the first one is A. no matter how the arrangements are, each person, including A, has to have a seat somewhere around the table. Then we just start with A EVERY SINGLE TIME. So A's position is fixed. What is changing is the other 4 people's seats relative to A.
10#
发表于 2011-6-6 22:57:39 | 只看该作者
h(100) = 2*4*6*8*. . .*94*96*98*100 = 2^50 *(1*2*3*4* . . . *47*48*49*50)

Apparently, h(100) have all the prime factors between 2 and 47.

h(100) and h(100) + 1 are two consecutive integers. As such, these two numbers are co-prime, meaning they do not share prime factors.

Therefore, h(100) + 1 do not have prime factors between 2 and 47. The smallest prime factor it could have is 53.
-- by 会员 sdcar2010 (2011/4/2 12:55:29)



可不可以说一下具体53怎么算出来的咧,前面都懂了。谢谢!
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