求助几道PREP数学题，谢谢

yifeir · 发表于 2008-8-3 21:29:00

1 905-!-item-!-187;#058&000575

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

答案是E，不明白怎么求出来的，难道代数？

2 2463-!-item-!-187;#058&002761

For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6 ?

(A) One

(B) Two

(C) Three

(D) Four

(E) Five

答案是C 没有读懂题目，希望能帮忙解释下，谢谢。

3. 563-!-item-!-187;#058&000364

If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ?

(1) n = 2

(2) m = 1

答案是B，不选C么？为什么不用知道n呢？

谢谢大家。

For a finite sequence of nonzero numbers, the number of variations in sign is defined as the number of pairs of consecutive terms of the sequence for which the product of the two consecutive terms is negative. What is the number of variations in sign for the sequence 1, -3, 2, 5, -4, -6 ?

(A) One

(B) Two

(C) Three

(D) Four

(E) Five

答案是C 没有读懂题目，希望能帮忙解释下，谢谢。

3. 563-!-item-!-187;#058&000364

If n and m are positive integers, what is the remainder when 3^(4n + 2 + m) is divided by 10 ?

(1) n = 2

(2) m = 1

答案是B，不选C么？为什么不用知道n呢？

谢谢大家。

[此贴子已经被作者于2008-8-3 21:30:27编辑过]

yifeir · 发表于 2008-8-3 21:31:00

不好意思，后两道题和前面重复了，可是我编辑好多次都消除不下去，对不起啊。给帮忙的人添麻烦了。哎。

杰西卡的戈多 · 发表于 2008-8-3 22:27:00

（1）这个我只能推导到h(100)+1=100!+1,后面就不会了，感觉这个让算质因子会很麻烦啊

（2）非0的有限序列，符号变更的次数取决于连续数列中两个连续的数的符号是不是否定的。这个题里面，1，-3是一次，-3，2是一次，5，-4是一次，也就是说总共3次

（3）不懂……

完了，这三题真难啊，关注此贴~

yifeir · 发表于 2008-8-3 23:19:00

以下是引用杰西卡的戈多在2008-8-3 22:27:00的发言：

（1）这个我只能推导到h(100)+1=100!+1,后面就不会了，感觉这个让算质因子会很麻烦啊

（2）非0的有限序列，符号变更的次数取决于连续数列中两个连续的数的符号是不是否定的。这个题里面，1，-3是一次，-3，2是一次，5，-4是一次，也就是说总共3次

（3）不懂……

完了，这三题真难啊，关注此贴~

我怎么理解第2题的意思是取决于连续两个数的乘积是不是负数呢？不过大概我明白了。呵呵。谢谢你。

怎么没有其他人来看看呢？再次谢谢大家了。

愁予 · 发表于 2008-8-3 23:31:00

(3)任何一个数被10除余数就是他的尾数，所以只要知道3^(4n + 2 + m)的尾数就可以了。3的n次方的尾数循环式3，9，7，1。4个数一个循环，所以由（2）3^(4n+3)可知这个数的尾数是7，余数就是7

ricken_ivy · 发表于 2008-8-4 01:21:00

以下是引用愁予在2008-8-3 23:31:00的发言：
(3)任何一个数被10除余数就是他的尾数，所以只要知道3^(4n + 2 + m)的尾数就可以了。3的n次方的尾数循环式3，9，7，1。4个数一个循环，所以由（2）3^(4n+3)可知这个数的尾数是7，余数就是7

谢谢你的解释啊我之前也做错了，后来就是采取带入法看他们的余数是不是一个CYCLE。果然如此。我发现DS这个考点的题目都可以考虑用带入法来破解，一个是时间上快而且准确率比较高。有时候我们看似是E其实仔细一计算发现答案都是唯一的。大家一定要把题目里面的那些条件读仔细了。

jh1904 · 发表于 2008-8-4 09:48:00

1. has been asked many times here.

h(100) = 2*4*6...*100 = 2^50 * 50!, so h(100) has factors from 1 to 50 including all the prime numbers from 2 to 47.

As h(100)+1 divided by any factors of h(100) will leave a remainder 1, any factor of h(100)+1 other than 1 must be greater than 50.

So the smallest prime factor of h(100)+1 must be greater than 50.

2. yifeir is right. 相邻两数乘积为负。

yifeir · 发表于 2008-8-4 14:47:00

以下是引用jh1904在2008-8-4 9:48:00的发言：
1. has been asked many times here.

h(100) = 2*4*6...*100 = 2^50 * 50!, so h(100) has factors from 1 to 50 including all the prime numbers from 2 to 47.

As h(100)+1 divided by any factors of h(100) will leave a remainder 1, any factor of h(100)+1 other than 1 must be greater than 50.

So the smallest prime factor of h(100)+1 must be greater than 50.

2. yifeir is right. 相邻两数乘积为负。

谢谢你。可是第一题这句话我还是没有看懂，

As h(100)+1 divided by any factors of h(100) will leave a remainder 1, any factor of h(100)+1 other than 1 must be greater than 50.

为什么如果h(100)+1能被所有的h(100)因数除余1，就会得到了它的因数大于50？

桃儿 · 发表于 2008-8-4 15:28:00

我也是，第二题勉强懂了，1，3题还是没明白！

january_ye · 发表于 2008-8-4 15:34:00

http://forum.chasedream.com/dispbbs.asp?BoardID=22&ID=124700

前人第一题的讨论链接，比较清楚，涉及到互质等很多知识点

求助几道PREP数学题，谢谢

求助几道PREP数学题，谢谢

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