2道难题

一叶知秋

67.Five boxes maximum capacity currently hold pieces
90 70
80 70
50 30
80 50
40 40
rearrange it so that the maximum capacity - holding pieces for all five are
the same. So second box should be ?
[确认]64
[思路]所有差额的和为：20+10+20+30+0=80，5对数每对的平均差额为：80/5=16
因此second box：80－16＝64
答案好象应为70+16=86，班主是否搞错了？
32. (PS)1-99,inclusive,其product为(n+1)n的multiple是几分之几？
[确认]2/3
此题不知如何解，求助。

brigittew

i rememebered the answer of the question should be:
first, account the number of n, 99/3=33
then, the number of n+1, (since n has 99 possiblity, n+1 also has 99)99/3=33
so the posssibility of the multiple of 3 here is (33+33)/99=2/3
i dont know whether the answer is calculate as this.

skycrash

liu9001 why you say
such as :
2=2*1
6=3*2
12=4*3
...
72=9*8
在往上就超过99了
90=10*10 is also fit for the requirement.
Totally have 9 case.
1*2, 4*5, 7*8 can't be divided by 3.
so 9-3/9=2/3
are you with me?

tongxun

cook：
32. (PS)1-99,inclusive,其product为(n+1)n的multiple是几分之几？
[确认]2/3
----------------
准确的旧机经是这样的：
n is a whole number in 1-99，inclusive。what is the possibility of the product of n（n+1） is mutiples of 3？

cook

Can u do more explanation? Thank , tongxun!

tongxun

32、1-99 是限定N 的。
112、这个题目不完整

cook

112.K*=K^2+1,K=?  K* (1)K=4n+1 (2)18<=K<=36 [确认]A ??? ========================= How to do? I can't understand. 32. the question has some problem, I think. " (其product)为(n+1)n的multiple是几分之几"

smilingme

咦，好像是哎！
容我再想想啊~~~~~

不好意思 ：）

liu9001

smilingme

第二题是如何得到的？
按我的理解：
1-99内只有2，6，12，20，30，42，56，72可以为n(n+1)
such as :
2=2*1
6=3*2
12=4*3
...
72=9*8
在往上就超过99了

smilingme

1.没错。
  问的是第二个盒子嘛。第二个盒子原来是80啊，要使得五个盒子差额相等，就要用原来的减去这个差额啊，得到现在第二个盒子里面的数量。你觉得有什么地方不对么？
  你的这个式子是怎么得到的呢？

2.66/99=2/3