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prep1中的另外几个数学题???

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楼主
发表于 2008-8-9 02:54:00 | 只看该作者

prep1中的另外几个数学题???

这些题目看上去很简单,我就是弄错了。跪求

1.

For how many integers n is 2^n = n^2 ?

(A) None

(B) One

(C) Two

(D) Three

(E) More than three

2

A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10 percent greater than the population of any other district.  What is the minimum possible population that the least populated district could have?

(A) 10,700

(B) 10,800

(C) 10,900

(D) 11,000

(E) 11,100

3.

38, 69, 22, 73, 31, 47, 13, 82

Which of the following numbers is greater than three-fourths of the numbers but less than one-fourth of the numbers in the list above?

(A) 56

(B) 68

(C) 69

(D) 71

(E) 73

4.

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


[此贴子已经被作者于2008-8-9 3:02:56编辑过]
沙发
发表于 2008-8-9 12:21:00 | 只看该作者

好多题啊……

拜托以后贴题时给个答案啊,那样比较容易想思考过程。

第四题前两天见到了,你搜一下吧。


[此贴子已经被作者于2008-8-9 14:17:41编辑过]
板凳
发表于 2008-8-9 12:24:00 | 只看该作者

看错题目了……


[此贴子已经被作者于2008-8-9 14:15:50编辑过]
地板
发表于 2008-8-9 13:18:00 | 只看该作者

第一个,不会,试试,好像只有2和4,所以是C答案。

第二个,是A+A*1。1*10=13200,000,那么计算A,为D。(把一个的10%平分给其它10,这样差别最大)

第三个,把八个数字排列,前3/4和后1/4中间的数字是,D答案,这个原来有人贴过的。

第四个,不会,那位大师可以给讲讲。

已改正,谢谢。


[此贴子已经被作者于2008-8-10 12:18:33编辑过]
5#
 楼主| 发表于 2008-8-10 04:43:00 | 只看该作者

1。 第一题的答案是C,我就感到这种题作的时候是不是依赖自己的常识,万一我当时想不起来4的平方也等于16就完了,这种题有没有其他的解法??

2.第二题的答案是D。quietliu的回答是不是把这个当成了下面的第三个题目?

3.第三题答案是71,我感觉这个题目就是小孩玩过加加一样。搞得我当时就把最小的和最大的数字进行4分之一和四分之三,然后在中间找,又上了当了。

第四个题目我也在思考。答案是E。


[此贴子已经被作者于2008-8-10 4:53:39编辑过]
6#
发表于 2010-3-7 16:12:44 | 只看该作者
第四题的解答,从ManhattanGMAT抄来的:

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.

非常聪明的解答~
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