请教大家几道题..

hedgeforfu

请大家不要鄙视我...55...关于余数和整除的问题一直困扰我....

8. 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
这题选B
11.If p is a positive odd integer, what is the remainder when p is divided by 4 ?
(1) When p is divided by 8, the remainder is 5.
(2) p is the sum of the squares of two positive integers.
选D
35.  If p and n are positive integers and p > n, what is the remainder when p2 - n2 is divided by 15 ?
(1) The remainder when p + n is divided by 5 is 1.
The remainder when p - n is divided by 3 is 1.
选E
还有下面这道题为何不选D..而要选C。。
49Does x + y = 5 ?
(1) 4x + y = 17
（2）x + 4y = 8
选C

23. At least 100 students at a certain high school study Japanese. If 4 percent of the students at the school who study French also study Japanese, do more students at the school study French than Japanese?

(1) 16 students at the school study both French and Japanese

. (2) 10 percent of the students at the school who study Japanese also study French.

题意没读懂，这俩已知条件有啥用....不都是说的同时学习日语法语么..\
这题选B

onedayling

8. 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
这题选B

m=1的时候原式=3^(4n+3)=27x(3^4n)
此时无论n的值是多少余数一定是7
因为27/10=2……7

franker

11.
(1)p=8n+5
p/4=(2n+1)……1
(2)p=axa+bxb并且是奇数
所以a和b必定一个是奇数一个是偶数
偶数的平方肯定能被4整除
而奇数可以表示成2c+1
所以奇数的平方=4cxc+4c+1
除以4的余数肯定是1

选D

franker

35.
(1)p+n=5a+1
不知道p-n情况，没用
(2)p-n=3b+1
不知道p+n，还是白说
两个条件联立
p2-n2=15ab+5a+3b+1
5a+3b+1是个什么东西就很难说了
so~E

nieyuhang

49有两个未知数诶
没有两个方程联立怎么可能有确定解啊

23咱假设学习日语的a人，学习法语的b人，两种同时学的c人，问b是否大于a
题目给出条件a+b-c>100,4%b=c,即b=25c
(1)16%(a+b-c)=c
化简a+b-c=6.25c
即a+b=7.25c
难以比较
(2)10%a=c
化简a=10c
因为c>0所以a<b

hedgeforfu

非常感谢这位大牛...49我的问题是...。为什么不可以选D，x+y=5和条件1,2 分别联立，能解出有效的解即是能推出。

nieyuhang

你题目理解错啦
49是问你如果已经知道（1）或者（2），x+y是不是等于5

hedgeforfu

哎呀....不好意思O(∩_∩)O         多谢多谢~~~~