两题数学求解

昨天我发帖问了，但是还是不明白，请好心人帮忙解答一下吧，下个星期而就要考啦。

1.Amy's grade was the 90th percentile of the 80 grades for her class. Of the 100 grades from another class, 19 were higher than Amy's, and the rest were lower. If no other grade was the same as Amy's grade, then Amy's grade was what percentile of the grades of the two classes combined?

这题答案为85th. 我主要是不理解"percentile",我查了一下是说高于总分百分之几的分数，怎么理解啊？

2.DS 题

The intergers m and p are such that 2<m<p and m is not a factor of p. If r is the reminder when p is divided by m, is r>1?

（1）the greatest common factor of m and p is 2.

(2) the least common mutiple of m and p is 30

这题选A。 但是（2）不行吗？ 还有这题具体解法是什么啊？

谢谢啦

没人理啊——

还有一题。

If n is a positive integer and r is the remainder when (n – 1)(n + 1) is divided by 24, what is the value of r?

(1)     2 is not a factor of n.

(2)     3 is not a factor of n.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.

B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.

C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.

D. EACH statement ALONE is sufficient.

E. Statements (1) and (2) TOGETHER are NOT sufficient.

我想问像这种有关remainder的可以为0吗？

还有这题的解法谢谢

先说正经的

第二题

（1）m和p的最大公约数是2，那么他俩都是偶数，二者之差一定大于1，故条件充分

（2）最小公倍数是30，又2<m<p，30=2×3×5。举例：余数>1的情况，m=6，p=15；余数<1的情况，m=5，p=6.故不成立

所以此题选a

第三题

这题选c

先说两个条件同时成立时为什么对。(1)  
                2 is not a factor of n 说明n是奇数，则(n – 1)(n + 1)相当于两个相邻的偶数之积，两个相邻的偶数一定一个是2的倍数，另一个是4的倍数，所以(n – 1)(n + 1)里含有8这个因子。(2)     3 is not a factor of n 。n不是三的倍数，则(n – 1)或(n + 1)必然有一个是3的倍数，连续的三个正整数必然有一个是3的倍数，所以(n – 1)(n + 1)又一定含有3这个因子。综上，(n – 1)(n + 1)既能被8又能被3整除，故一定能被24整除，remainder r等于零。所以两条件结合，成立。

单独看（1），不充分,举反例：n=3，(n – 1)(n + 1)=8，此时r=8；n=5，(n – 1)(n + 1)=24，r=0

（2），不充分：n=2，(n – 1)(n + 1)=3，r=3； n=5，(n – 1)(n + 1)=24，r=0

故此题选c

最后想说第一题