【数学讨论稿101-200】8.24-8.26数学讨论稿101-200(9.6 大量新增）

joycezbj

103 如果狗主翻译没错 一个数是另一个数的平方才叫perfect square 那么0和1是不是都应该剔除呢？因为结果为自身的平方 所以答案应该是2~100 有99个？？？
106 为什么不考虑(x+1)(x+p)的情况呢？这样算出来m=-(p+1)？？？
大神在哪里，求解~~~

kagomearisa

179里答案q怎么不见了。。。应该是（5x+4w+3q+2r+v）/5吧 最后一天的range s不用考虑，因为是求开盘均值

shen_Ali

112题 84还有一个因数是14，正好是选项B，所以答案还是能算出来的~

b_brother5

楼主辛苦了！加油！

joycezbj

112 应该是比运动员人数少4吧 如果剩余的饮料比运动员人数还多4 干嘛不每个人多分一瓶... 如果是少4的话求84的因数 狗主有没有可能把E的42记成32了呢 哈哈哈 这样就只有17不可能啦

hqq94

李睿 发表于 2015-8-26 11:10
辛苦了楼主

谢谢！一起加油

shen_Ali

前途迷茫 发表于 2015-8-25 07:08
麻烦楼主再看下117题。 我的理解是box 里原来有78个糖，分完以后还剩6个，那么就是分了72个。 然后盘子里的 ...

我也是这么想的哎，等楼主回复~

yanjingxiaofei

数讨君~我做manhanttan模考也做到sibling这题了，我看到更新的讨论稿加了一个网友给的版本，下面manhanttan给的答案解释是另一种方法，你看看，也可以加上呀~

题目：In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

解释：We are told that 4 people have exactly 1 sibling. This would account for 2 sibling relationships (e.g. AB and CD). We are also told that 3 people have exactly 2 siblings. This would account for another 3 sibling relationships (e.g. EF, EG, and FG). Thus, there are 5 total sibling relationships in the group.

Additionally, there are (7 x 6)/2 = 21 different ways to chose two people from the room.

Therefore, the probability that any 2 individuals in the group are siblings is 5/21. The probability that any 2 individuals in the group are NOT siblings = 1 – 5/21 = 16/21.

The correct answer is E.

棋梓

Virginie 发表于 2015-8-25 22:14
M应该是4，因为4是唯一一个2+2或者2*2都可以的。。。你可以想想九九乘法表。。。
如果从一开始算的话，1和 ...

啊啊啊～我要抓狂有没有！！！人家求百位数，我一直在琢磨个位...我把自己的膝盖收了吧
谢谢小v妹纸～

zengymtina

162感觉题目是说第一年存的不拿回，然后再加上第二年存进去的，到第七年一共有多少钱？