[求助]天山-3-32

smalltree

up, up!

NN help ....how to sovle such kind of question? Xianxieguole!!!

windlake

以下是引用xulijun在2005-2-19 20:57:00的发言：

条件1:x=12m+5  x=8m+4m+5   当m=1(奇数),余数为1,当m=2(偶数)时,m=5

条件2:x=18n+7 x=16n+2n+7   当n=1时,余数为1,当n=2时，余3，n＝3，余5

（直接改了，xulijun请别介意）

所以条件１和条件２可以推出，余数为１。5

这个方法是万能的，我就是用这个方法算的，可惜xulijun没解完，

条件2继续下去，当n＝3时，余数为5，

条件1与条件2的交集是1，5，不唯一，故而选E

这个方法可以解所有的这类题，36m＋n的说法有许多不足，因为36的因子有2和3，而8与3是没有交集的，这种方法更适用于单一因子，比如，被8m＋n，被2。4除，都余n。

讨论请继续！

WalkByFaith

To windlake:

you haven't finished it, either.

Q32:
What is the remainder when the positive integer x is divided by 8?
(1) When x is divided by 12, the remainder is 5.
(2) When x is divided by 18, the remainder is 7.

条件1:x=12m+5  x=8m+4m+5   当m=1(奇数),余数为1,当m=2(偶数)时,m=5

条件2:x=18n+7 x=16n+2n+7   当n=1时,余数为1,当n=2时，余3，n＝3，余5

when n=4, 余7

所以条件１和条件２可以推出，余数为 1, 3.

Furthermore, in TS, this item should be:

Q32:
What is the remainder when the positive integer x is divided by 8?
(1) When x is divided by 12, the remainder is 5.
(2) When x is divided by 18, the remainder is 11

therefore, by using your method:

条件1:x=12m+5  x=8m+4m+5   当m=1(奇数),余数为1,当m=2(偶数)时,m=5

条件2:y=18n+11 y=16n+2n+11   当n=1时,余数为5,当n=2时，余7，n＝3，余1, when n=4, 余3,when n=5, 余5, when n=6, 余7,when n=7, 余1, when n=8, 余3 so on and so forth

therefore,

条件1与条件2的交集是1,5，not唯一，故而选e. Great idea, though.

open to discuss....

WalkByFaith

if 条件1与条件2的交集是5，say, 唯一，will you choose C?

OR,

we need to confirm whether X is the same number?

otherwise, still can not satisfy both of them, right?

riverhouse

cicila is good!

爱无伤

明白了,谢谢~

hhjc

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